Derivative Practice Problems Answers

CALCULUS I Solutions to Practice Problems

Derivatives

Paul Dawkins

Calculus I

Table of Contents Preface ............................................................................................... Error! Bookmark not defined. Derivatives ......................................................................................... Error! Bookmark not defined.

The Definition of the Derivative ............................................................. Error! Bookmark not defined. Interpretations of the Derivative ........................................................... Error! Bookmark not defined. Differentiation Formulas ........................................................................ Error! Bookmark not defined. Product and Quotient Rule ..................................................................... Error! Bookmark not defined. Derivatives of Trig Functions ................................................................. Error! Bookmark not defined. Derivatives of Exponential and Logarithm Functions .......................... Error! Bookmark not defined. Derivatives of Inverse Trig Functions .................................................... Error! Bookmark not defined. Derivatives of Hyperbolic Functions ...................................................... Error! Bookmark not defined. Chain Rule ................................................................................................ Error! Bookmark not defined. Implicit Differentiation ........................................................................... Error! Bookmark not defined. Related Rates ........................................................................................... Error! Bookmark not defined. Higher Order Derivatives ........................................................................ Error! Bookmark not defined. Logarithmic Differentiation .................................................................... Error! Bookmark not defined.

Preface Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more or less detail than other solutions. The level of detail in each solution will depend up on several issues. If the section is a review section, this mostly applies to problems in the first chapter, there will probably not be as much detail to the solutions given that the problems really should be review. As the difficulty level of the problems increases less detail will go into the basics of the solution under the assumption that if you’ve reached the level of working the harder problems then you will probably already understand the basics fairly well and won’t need all the explanation. This document was written with presentation on the web in mind. On the web most solutions are broken down into steps and many of the steps have hints. Each hint on the web is given as a popup however in this document they are listed prior to each step. Also, on the web each step can be viewed individually by clicking on links while in this document they are all showing. Also, there are liable to be some formatting parts in this document intended for help in generating the web pages that haven’t been removed here. These issues may make the solutions a little difficult to follow at times, but they should still be readable.

Derivatives

© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

Calculus I

The Definition of the Derivative 1. Use the definition of the derivative to find the derivative of, ( ) 6f x = Solution There really isn’t much to do for this problem other than to plug the function into the definition of the derivative and do a little algebra.

( ) ( ) ( )0 0 0 0

6 6 0lim lim lim lim 0 0h h h h

f x h f xf x

h h h→ → → →

+ − −′ = = = = =

So, the derivative for this function is,

( ) 0f x′ = 2. Use the definition of the derivative to find the derivative of, ( ) 3 14V t t= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( )0 0

3 14 3 14lim limh h

V t h V t t h tV t

h h→ →

+ − − + − −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some quick algebra and we’ll be done.

( ) ( )0 0 0

3 14 14 3 14 14lim lim lim 14 14h h h

t h t hV th h→ → →

− − − + −′ = = = − = −

The derivative for this function is then,

( ) 14V t′ = −

© 2007 Paul Dawkins 2 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

3. Use the definition of the derivative to find the derivative of, ( ) 2g x x= Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( )2 2

0 0lim limh h

g x h g x x h xg x

h h→ →

+ − + −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Step 2 Now all that we need to do is some quick algebra and we’ll be done.

( ) ( ) ( )2 2 2

0 0 0

22lim lim lim 2 2h h h

h x hx xh h xg x x h xh h→ → →

++ + −′ = = = + =


( ) 2g x x′ = 4. Use the definition of the derivative to find the derivative of, ( ) 210 5Q t t t= + − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( ) ( )2 2

0 0

10 5 10 5lim limh h

t h t h t tQ t h Q tQ t

h h→ →

+ + − + − + −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete.


Calculus I

Step 2 Now all that we need to do is some algebra (and it might get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( )

( ) ( )

2 2 2

0

0 0

10 5 5 2 10 5lim

5 2lim lim 5 2 5 2

h

h h

t h t th h t tQ th

h t ht h t

h

→

→ →

+ + − − − − − +′ =

− −= = − − = −


( ) 5 2Q t t′ = − 5. Use the definition of the derivative to find the derivative of, ( ) 24 9W z z z= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( ) ( )2 2

0 0

4 9 4 9lim limh h

z h z h z zW z h W zW z

h h→ →

+ − + − −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some algebra (and it might get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.


Calculus I

( ) ( )

( ) ( )

2 2 2

0

2 2 2

0

0 0

4 2 9 9 4 9lim

4 8 4 9 9 4 9lim

8 4 9lim lim 8 4 9 8 9

h

h

h h

z zh h z h z zW z

hz zh h z h z z

hh z h

z h zh

→

→

→ →

+ + − − − +′ =

+ + − − − +=

+ −= = + − = −


( ) 8 9W z z′ = − 6. Use the definition of the derivative to find the derivative of, ( ) 32 1f x x= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( )3 3

0 0

2 1 2 1lim limh h

x h xf x h f xf x

h h→ →

+ − − −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some algebra (and it might get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( ) ( )

( ) ( )

3 2 2 3 3

0

3 2 2 3 3

0

2 22 2 2

0 0

2 3 3 1 2 1lim

2 6 6 2 1 2 1lim

6 6 2lim lim 6 6 2 6

h

h

h h

x x h xh h xf x

hx x h xh h x

hh x xh h

x xh h xh

→

→

→ →

+ + + − − +′ =

+ + + − − +=

+ += = + + =


Calculus I


( ) 26f x x′ = 7. Use the definition of the derivative to find the derivative of, ( ) 3 22 1g x x x x= − + − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( ) ( ) ( )3 2 3 2

0 0

2 1 2 1lim limh h

x h x h x h x x xg x h g xg x

h h→ →

+ − + + + − − − + −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete. Step 2 Now all that we need to do is some algebra (and it will get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( ) ( ) ( )

( )

( )

3 2 2 3 2 2 3 2

0

3 2 2 3 2 2 3 2

0

2 2

0

2 2 2

0

3 3 2 2 1 2 1lim

3 3 2 4 2 1 2 1lim

3 3 4 2 1lim

lim 3 3 4 2 1 3 4 1

h

h

h

h

x x h xh h x xh h x h x x xg x

hx x h xh h x xh h x h x x x

hh x xh h x h

hx xh h x h x x

→

→

→

→

+ + + − + + + + − − − + −′ =

+ + + − − − + + − − + − +=

+ + − − +=

= + + − − + = − +


( ) 23 4 1g x x x′ = − +


Calculus I

8. Use the definition of the derivative to find the derivative of,

( ) 5R zz

=

Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( )0 0

1 5 5lim limh h

R z h R zR z

h h z h z→ →

+ − ′ = = − +

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also note that in order to make the problem a little easier to read rewrote the rational expression in the definition a little bit. This doesn’t need to be done, but will make things a little nicer to look at. Step 2 Next we need to combine the two rational expressions into a single rational expression.

( ) ( )( )0

5 51limh

z z hR z

h z z h→

− +′ = +

Step 3 Now all that we need to do is some algebra and we’ll be done.

( ) ( ) ( ) ( ) 20 0 0

1 5 5 5 1 5 5 5lim lim limh h h

z z h hR zh z z h h z z h z z h z→ → →

− − − −′ = = = = − + + +


( ) 2

5R zz

′ = −


( ) 14

tV tt+

=+

Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( )0 0

1 1 1lim lim4 4h h

V t h V t t h tV th h t h t→ →

+ − + + + ′ = = − + + +


Calculus I

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Also note that in order to make the problem a little easier to read rewrote the rational expression in the definition a little bit. This doesn’t need to be done, but will make things a little nicer to look at. Step 2 Next we need to combine the two rational expressions into a single rational expression.

( ) ( )( ) ( )( )( )( )0

1 4 1 41lim4 4h

t h t t t hV t

h t h t→

+ + + − + + +′ = + + +

Step 3 Now all that we need to do is some algebra (and it will get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

( ) ( )( )( )

( )( )

( )( ) ( )( ) ( )

2 2

0

2 2

0

20 0

5 4 4 5 41lim4 4

1 5 4 4 5 4lim4 4

1 3 3 3lim lim4 4 4 4 4

h

h

h h

t th t h t th t hV t

h t h t

t th t h t th t hh t h t

hh t h t t h t t

→

→

→ →

+ + + + − + + + +′ =

+ + + + + + + − − − − −

= + + +

= = = + + + + + + +


( )( )2

34

V tt

′ =+


( ) 3 4Z t t= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( )0 0

3 4 3 4lim limh h

t h tZ t h Z tZ t

h h→ →

+ − − −+ −′ = =


Calculus I

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Step 2 Next we need to rationalize the numerator.

( )( )( ) ( )( )

( )( )0

3 4 3 4 3 4 3 4lim

3 4 3 4h

t h t t h tZ t

h t h t→

+ − − − + − + −′ =

+ − + −


( ) ( ) ( )( )( ) ( )( )

( )( ) ( )

0 0

0 0

3 4 3 4 3 3 4 3 4lim lim3 4 3 4 3 4 3 4

3 3 3lim lim2 3 43 4 3 43 4 3 4

h h

h h

t h t t h tZ th t h t h t h t

htt h th t h t

→ →

→ →

+ − − − + − − +′ = =+ − + − + − + −

= = =−+ − + −+ − + −

Be careful when multiplying out the numerator here. It is easy to lose track of the minus sign (or parenthesis for that matter) on the second term. This is a very common mistake that students make. The derivative for this function is then,

( ) 32 3 4

Z tt

′ =−


( ) 1 9f x x= − Step 1 First we need to plug the function into the definition of the derivative.

( ) ( ) ( ) ( )0 0

1 9 1 9lim limh h

x h xf x h f xf x

h h→ →

− + − −+ −′ = =

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems. Step 2


Calculus I

Next we need to rationalize the numerator.

( )( )( ) ( )( )

( )( )0

1 9 1 9 1 9 1 9lim

1 9 1 9h

x h x x h xf x

h x h x→

− + − − − + + −′ =

− + + −


( ) ( ) ( )( )( ) ( )( )

( )( ) ( )

0 0

0 0

1 9 1 9 1 9 9 1 9lim lim1 9 1 9 1 9 1 9

9 9 9lim lim2 1 91 9 1 91 9 1 9

h h

h h

x h x x h xf xh x h x h x h x

hxx h xh x h x

→ →

→ →

− + − − − − − +′ = =− + + − − + + −

− − −= = =

−− + + −− + + −

Be careful when multiplying out the numerator here. It is easy to lose track of the minus sign (or parenthesis for that matter) on the second term. This is a very common mistake that students make. The derivative for this function is then,

( ) 92 1 9

f xx

−′ =−

Interpretations of the Derivative 1. Use the graph of the function, ( )f x , estimate the value of ( )f a′ for

(a) 2a = − (b) 3a =


Calculus I

Solution Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. (a) 2a = − Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point let’s first sketch in a tangent line at the point on the graph.

Step 2 The function is clearly decreasing here and so we know that the derivative at this point will be negative. Now, from this sketch of the tangent line it looks like if we run over 1 we go down 4 and so we can estimate that,

( )2 4f ′ − = −

(b) 3a =


Calculus I

Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. Let’s first sketch in a tangent line at the point.

Step 2 The function is clearly increasing here and so we know that the derivative at this point will be positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 2 and so we can estimate that,

( )3 2f ′ =

2. Use the graph of the function, ( )f x , estimate the value of ( )f a′ for

(a) 1a = (b) 4a =

Solution


Calculus I

Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. (a) 1a = Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point let’s first sketch in a tangent line at the point on the graph.

Step 2 The function is clearly decreasing here and so we know that the derivative at this point will be positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 1 and so we can estimate that,

( )1 1f ′ =

(b) 4a = Step 1 Given that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. Let’s first sketch in a tangent line at the point.


Calculus I

Step 2 The function is clearly decreasing here and so we know that the derivative at this point will be positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 5 and so we can estimate that,

( )4 5f ′ =

3. Sketch the graph of a function that satisfies ( )1 3f = , ( )1 1f ′ = , ( )4 5f = , ( )4 2f ′ = − . Solution Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. Step 1 First, recall that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. So, let’s start off with a graph that has the given points on it and a sketch of a tangent line at the points whose slope is the value of the derivative at the points.


Calculus I

Step 2 Now, all that we need to do is sketch in a graph that goes through the indicated points and at the same time it must be parallel to the tangents that we sketched. There are many possible sketches that we can make here and so don’t worry if your sketch is not the same as the one here. This is just one possible sketch that meets the given conditions.

While, it’s not really needed here is a sketch of the function without all the extra bits that we put in to help with the sketch.


Calculus I

4. Sketch the graph of a function that satisfies ( )3 5f − = , ( )3 2f ′ − = − , ( )1 2f = , ( )1 0f ′ = ,

( )4 2f = − , ( )4 3f ′ = − . Solution Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to the function. Step 1 First, recall that one of the interpretations of the derivative is that it is the slope of the tangent line to the function at a particular point. So, let’s start off with a graph that has the given points on it and a sketch of a tangent line at the points whose slope is the value of the derivative at the points.

Step 2


Calculus I

Now, all that we need to do is sketch in a graph that goes through the indicated points and at the same time it must be parallel to the tangents that we sketched. There are many possible sketches that we can make here and so don’t worry if your sketch is not the same as the one here. This is just one possible sketch that meets the given conditions.

While, it’s not really needed here is a sketch of the function without all the extra bits that we put in to help with the sketch.

5. Below is the graph of some function, ( )f x . Use this to sketch the graph of the derivative,

( )f x′ .


Calculus I

Solution Hint 1 : Where will the derivative be zero? This gives us a couple of starting points for our sketch. Step 1 From the graph of the function it is pretty clear that we will have horizontal tangent lines at

2x = − , 1x = and 5x = . Because we will have horizontal tangents here we also know that the derivative at these points must be zero. Therefore, we know the following derivative evaluations. ( ) ( ) ( )2 0 1 0 5 0f f f′ ′ ′− = = = Hint 2 : Recall that the derivative can also be used to tell us where the function is increasing and decreasing. Knowing this we can use the graph to determine where the derivative will be positive and where it will be negative. Step 2 The points we found above break the x-axis up into regions where the function is increasing and decreasing. Recall that if the derivative is positive then the function is increasing and likewise if the derivative is negative then the function is decreasing. Using these ideas we can easily identify the sign of the derivative on each of the regions. Doing this gives,

( )( )( )( )

2 0

2 1 0

1 5 0

5 0

x f x

x f x

x f x

x f x

′< − >

′− < < <

′< < >

′> <

Hint 3 : At this point all we have to do is try and put all this together and come up with a sketch of the derivative.


Calculus I

Step 3 This is the tricky part of this problem. In the range 2x < − we know the derivative must be positive and that it must be zero at 2x = − so it makes sense that just to the left of 2x = − the derivative must be decreasing. In the range 2 1x− < < we know that the derivative will be negative and that it will be zero at the endpoints of the range. So, to the right of 2x = − the derivative will have to be decreasing (goes from zero to a negative number). Likewise, to the left of 1x = the derivative will have to be increasing (goes from a negative number to zero). Note that we don’t really know just how the derivative will behave everywhere in this range, but we can use the general behavior near the endpoints and go with the simplest way to connect the two up to get an idea of what the derivative should look like. Following similar reasoning we can see that the derivative should be increasing just to the right of

1x = (goes from zero to a positive number), decreasing just to the left of 5x = (goes from a positive number to zero) and decreasing just to the right of 5x = (goes from zero to a negative number). Step 4 So, putting all of this together here is a sketch of the derivative. Note that we included the a scale on the vertical axis if you would like to try and estimate some specific values of the derivative as we did in Example 4 of this section.


Calculus I

6. Below is the graph of some function, ( )f x . Use this to sketch the graph of the derivative,

( )f x′ .

Solution Hint 1 : Because the derivative of a function is also the slope of the tangent line. We can therefore determine actual values of the derivative at almost every spot. Step 1 Because the three portions of the function are actually lines and the tangent line to a line would just be the line itself we can easily compute the derivative on each portion of the curve. On each of the portions we can use the grid included on the graph to compute the slope of each part. Knowing the slope of the graph on each portion will in turn tell us the slope of the tangent line for each portion. This in turn tells us that the derivative on each of the three portions is then,

( )

( )

( )

1 121 23

2 2

x f x

x f x

x f x

′< − = −

′− < < =

′> = −

Hint 2 : What is the derivative at the “sharp points”? Step 2 Recall Example 4 from the previous section. In that example we showed that the derivative of the absolute value function does not exist at 0x = . The limit on the left side of 0x = (which gives the slope of the line on the left) and the limit on the right side of 0x = (which gives the slope of the line on the right) were different and so the overall limit did not exist. This in turn tells us that the derivative doesn’t exist at that point.


Calculus I

Here we have the same problem. We’ll leave it to you to verify that the right and left handed limits at 1x = − are not the same and so the derivative does not exist at 1x = − . Likewise, the derivative does not exist at 2x = . There will therefore be open dots on the graph at these two points. Step 3 Here is the sketch of the derivative of this function.

7. Answer the following questions about the function ( ) 24 9W z z z= − .

(a) Is the function increasing or decreasing at 1z = − ? (b) Is the function increasing or decreasing at 2z = ? (c) Does the function ever stop changing? If yes, at what value(s) of z does the function stop changing? (a) Is the function increasing or decreasing at 1z = − ? We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 5 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. So, from our previous work we know that the derivative is, ( ) 8 9W z z′ = − Now all that we need to do is to compute : ( )1 17W ′ − = − . This is negative and so we know that

the function must be decreasing at 1z = − .


Calculus I

(b) Is the function increasing or decreasing at 2z = ? Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here. The evaluation is : ( )2 7W ′ = . This is positive and so we know that the function must be

increasing at 2z = . (c) Does the function ever stop changing? If yes, at what value(s) of z does the function stop changing? Here all that we’re really asking is if the derivative is ever zero. So we need to solve,

( ) 90 8 9 08

W z z z′ = → − = ⇒ =

So, the function will stop changing at 98

z = .

8. What is the equation of the tangent line to ( ) 3 14f x x= − at 8x = .

Solution We know that the derivative of a function gives us the slope of the tangent line and so we’ll first need the derivative of this function. We computed this derivative in Problem 2 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. Note that we did use a different set of letters in the previous problem, but the work is identical. So, from our previous work (with a corresponding change of variables) we know that the derivative is, ( ) 14f x′ = − This tells us that the slope of the tangent line at 8x = is then : ( )8 14m f ′= = − . We also know

that a point on the tangent line is : ( )( ) ( )8, 8 8, 109f = − .

The tangent line is then, ( )109 14 8 3 14y x x= − − − = −


Calculus I

Note that, in this case the tangent is the same as the function. This should not be surprising however as the function is a line and so any tangent line (i.e. parallel line) will in fact be the same as the line itself.

9. The position of an object at any time t is given by ( ) 14

ts tt+

=+

.

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop moving? If yes, at what time(s) does the object stop moving? (a) Determine the velocity of the object at any time t. We know that the derivative of a function gives is the velocity of the object and so we’ll first need the derivative of this function. We computed this derivative in Problem 9 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. Note that we did use a different letter for the function in the previous problem, but the work is identical. So, from our previous work we know that the derivative is,

( )( )2

34

s tt

′ =+

(b) Does the object ever stop moving? If yes, at what time(s) does the object stop moving? We know that the object will stop moving if the velocity (i.e. the derivative) is zero. In this case the derivative is a rational expression and clearly the numerator will never be zero. Therefore, the derivative will not be zero and therefore the object never stops moving.

10. What is the equation of the tangent line to ( ) 5f xx

= at 12

x = ?

Solution We know that the derivative of a function gives us the slope of the tangent line and so we’ll first need the derivative of this function. We computed this derivative in Problem 8 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding.


Calculus I

Note that we did use a different set of letters in the previous problem, but the work is identical. So, from our previous work (with a corresponding change of variables) we know that the derivative is,

( ) 2

5f xx

′ = −

This tells us that the slope of the tangent line at 12

x = is then : 1 202

m f ′= = −

. We also

know that a point on the tangent line is : 1 1 1, ,102 2 2

f = .

The tangent line is then,

110 20 20 202

y x x = − − = −

11. Determine where, if anywhere, the function ( ) 3 22 1g x x x x= − + − stops changing.

Solution We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 7 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. From our previous work (with a corresponding change of variables) we know that the derivative is, ( ) 23 4 1g x x x′ = − + If the function stops changing at a point then the derivative will be zero at that point. So, to determine if we function stops changing we will need to solve,

( )

( )( )

2

0

3 4 1 013 1 1 0 , 13

g x

x x

x x x x

′ =

− + =

− − = ⇒ = =

So, the function will stop changing at 13

x = and 1x = .


Calculus I

12. Determine if the function ( ) 3 4Z t t= − increasing or decreasing at the given points.

(a) 5t = (b) 10t = (c) 300t = (a) 5t = We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 10 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. So, from our previous work we know that the derivative is,

( ) 32 3 4

Z tt

′ =−

Now all that we need to do is to compute : ( ) 352 11

Z ′ = . This is positive and so we know that

the function must be increasing at 5t = . (b) 10t = Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.

The evaluation is : ( ) 3102 26

Z ′ = . This is positive and so we know that the function must be

increasing at 10t = . (c) 300t = Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.

The evaluation is : ( ) 33002 896

Z ′ = . This is positive and so we know that the function must

be increasing at 300t = . Final Note


Calculus I

As a final note to all the parts of this problem let’s notice that we did not really need to do any evaluations. Because we know that square roots will always be positive it is clear that the derivative will always be positive regardless of the value of t we plug in. 13. Suppose that the volume of water in a tank for 0 6t≤ ≤ is given by ( ) 210 5Q t t t= + − .

(a) Is the volume of water increasing or decreasing at 0t = ? (b) Is the volume of water increasing or decreasing at 6t = ? (c) Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing? (a) Is the volume of water increasing or decreasing at 0t = ? We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 4 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding. So, from our previous work we know that the derivative is, ( ) 5 2Q t t′ = − Now all that we need to do is to compute : ( )0 5Q′ = . This is positive and so we know that the

volume of water in the tank must be increasing at 0t = . (b) Is the volume of water increasing or decreasing at 6t = ? Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here. The evaluation is : ( )6 7Q′ = − . This is negative and so we know that the volume of water in the

tank must be decreasing at 6t = . (c) Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing? Here all that we’re really asking is if the derivative is ever zero. So we need to solve,

( ) 50 5 2 02

Q t t t′ = → − = ⇒ =


Calculus I

So, the volume of water will stop changing at 52

.

Differentiation Formulas 1. Find the derivative of ( ) 36 9 4f x x x= − + .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section.

( ) 218 9f x x′ = − 2. Find the derivative of 4 22 10 13y t t t= − + . Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section.

38 20 13dy t tdt

= − +

3. Find the derivative of ( ) 7 74 3 9g z z z z−= − + .


( ) 6 828 21 9g z z z−′ = + +


Calculus I

4. Find the derivative of ( ) 4 3 29 8 12h y y y y− − −= − + + .


( ) 5 4 34 27 16h y y y y− − −′ = − + −

5. Find the derivative of 3 48 2y x x x= + − . Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to convert the roots to fractional exponents before you start taking the derivative. Here is the rewritten function.

11 132 48 2y x x x= + −

The derivative is,

21 332 41 8 1

2 3 2dy x x xdx

−− −= + −

6. Find the derivative of ( ) 5 33 7 810 6 3f x x x x= − + − .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to convert the roots to fractional exponents before you start taking the derivative. Here is the rewritten function.

( ) ( ) ( ) ( )3 871 1 1

3 7 8 5 325 2 310 6 3 10 6 3f x x x x x x x= − + − = − + −


Calculus I

The derivative is,

( )2 5 2 55 55 3 5 32 23 7 8 710 6 6 16

5 2 3 2f x x x x x x x

− − ′ = − + = − +

7. Find the derivative of ( ) 3 5

4 1 86

f tt t t

= − + .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to rewrite the terms so that each of the t’s are in the numerator with negative exponents before taking the derivative. Here is the rewritten function.

( ) 1 3 514 86

f t t t t− − −= − +

The derivative is,

( ) 2 4 614 402

f t t t t− − −′ = − + −

8. Find the derivative of ( ) 4 103

6 1 18 3

R zz zz

= + − .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that you’ll need to rewrite the terms so that each of the z’s are in the numerator with negative exponents and rewrite the root as a fractional exponent before taking the derivative. Here is the rewritten function.

( )3

4 102 1 168 3

R z z z z− − −= + −

The derivative is,


Calculus I

( ) ( ) ( )5 5

5 11 5 112 23 1 1 1 106 4 10 92 8 3 2 3

R z z z z z z z− −− − − − ′ = − + − − − = − − +

9. Find the derivative of ( )23 9z x x= − .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to multiply the function out before we take the derivative. Here is the rewritten function.

33 9z x x= − The derivative is,

29 9dz xdx

= −

10. Find the derivative of ( ) ( )( )24 2g y y y y= − + .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to multiply the function out before we take the derivative. Here is the rewritten function. ( ) 3 22 8g y y y y= − − The derivative is,

( ) 23 4 8g y y y′ = − −


Calculus I

11. Find the derivative of ( )34 7 8x xh x

x− +

= .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to divide the function out and simplify before we take the derivative. Here is the rewritten function.

( )3

2 14 7 8 4 7 8x xh x x xx x x

−= − + = − +

The derivative is,

( ) 28 8h x x x−′ = −

12. Find the derivative of ( )5 3

3

5 2y y yf yy

− += .

Solution There isn’t much to do here other than take the derivative using the rules we discussed in this section. Remember that in order to do this derivative we’ll first need to divide the function out and simplify before we take the derivative. Here is the rewritten function.

( )5 3

2 23 3 3

5 2 5 2y y yf y y yy y y

−= − + = − +

The derivative is,

( ) 32 4f y y y−′ = − 13. Determine where, if anywhere, the function ( ) 3 29 48 2f x x x x= + − + is not changing.

Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to do this problem.


Calculus I

Solution Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. So, the function won’t be changing if its rate of change is zero and so all we need to do is find the derivative and set it equal to zero to determine where the rate of change is zero and hence the function will not be changing. First the derivative, and we’ll do a little factoring while we are at it.

( ) ( ) ( )( )2 23 18 48 3 6 16 3 8 2f x x x x x x x′ = + − = + − = + − Step 2 Now all that we need to do is set this equation to zero and solve. ( ) ( )( )0 3 8 2 0f x x x′ = ⇒ + − = We can easily see from this that the derivative will be zero at 8x = − and 2x = . The function therefore not be changing at,

8 and 2x x= − = 14. Determine where, if anywhere, the function 4 3 22 3y z z z= − − is not changing. Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to do this problem. Solution Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. So, the function won’t be changing if its rate of change is zero and so all we need to do is find the derivative and set it equal to zero to determine where the rate of change is zero and hence the function will not be changing. First the derivative, and we’ll do a little factoring while we are at it.

( )3 2 28 3 6 8 3 6dy z z z z z zdz

= − − = − −

Step 2


Calculus I

Now all that we need to do is set this equation to zero and solve.

( )2 2

0

8 3 6 0 0, 8 3 6 0

dydz

z z z z z z

=

− − = → = − − =

We can easily see from this that the derivative will be zero at 0z = , however, because the quadratic doesn’t factor we’ll need to use the quadratic formula to determine where, if anywhere, that will be zero.

( ) ( )( )

( )

23 3 4 8 6 3 2012 8 16

z± − − − ±

= =

The function therefore not be changing at,

3 201 3 2010 1.07359 0.6985916 16

z z z+ −= = = = = −

15. Find the tangent line to ( ) 16 4g x xx

= − at 4x = .

Hint : Recall the various interpretations of the derivative. One of them will help us do this problem. Solution Step 1 Recall that one of the interpretations of the derivative is that it gives slope of the tangent line to the graph of the function. So, we’ll need the derivative of the function. However before doing that we’ll need to do a little rewrite. Here is that work as well as the derivative.

( ) ( )1 1

1 22 22

16 216 4 16 2g x x x g x x xx x

−− −′= − ⇒ = − − = − −

Note that we rewrote the derivative back into rational expressions with roots to help with the evaluation. Step 2 Next we need to evaluate the function and derivative at 4x = .


Calculus I

( ) ( ) 2

16 16 24 4 4 4 4 24 4 4

g g′= − = − = − − −

Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )4 4 4 4 2 4 2 4y g g x x y x′= + − = − − − → = − + 16. Find the tangent line to ( ) 4 67 8 2f x x x x−= + + at 1x = − .

Hint : Recall the various interpretations of the derivative. One of them will help us do this problem. Solution Step 1 Recall that one of the interpretations of the derivative is that it gives slope of the tangent line to the graph of the function. So, we’ll need the derivative of the function.

( ) 3 7 37

4828 48 2 28 2f x x x xx

−′ = − + = − +

Note that we rewrote the derivative back into rational expressions help a little with the evaluation. Step 2 Next we need to evaluate the function and derivative at 1x = − .

( ) ( )1 7 8 2 13 1 28 48 2 22f f ′− = + − = − = − + + = Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )1 1 1 13 22 1 22 35y f f x x y x′= − + − + = + + → = + 17. The position of an object at any time t is given by ( ) 4 3 23 40 126 9s t t t t= − + − .

(a) Determine the velocity of the object at any time t. (b) Does the object ever stop changing? (c) When is the object moving to the right and when is the object moving to the left?


Calculus I

Solution Hint : Recall the various interpretations of the derivative. One of them is exactly what we need for this part. (a) Determine the velocity of the object at any time t. Recall that one of the interpretations of the derivative is that it gives the velocity of an object if we know the position function of the object. We’ve been given the position function of the object and so all we need to do is find its derivative and we’ll have the velocity of the object at any time t. The velocity of the object is then,

( ) ( )( )3 212 120 252 12 3 7s t t t t t t t′ = − + = − − Note that the derivative was factored for later parts and doesn’t really need to be done in general. Hint : If the object isn’t moving what is the velocity? (b) Does the object ever stop changing? The object will not be moving if the velocity is ever zero and so all we need to do is set the derivative equal to zero and solve. ( ) ( )( )0 12 3 7 0s t t t t′ = ⇒ − − = From this it is pretty easy to see that the derivative will be zero, and hence the object will not be moving, at,

0 3 7t t t= = = Hint : How does the direction (right vs. left) of movement relate to the sign (positive or negative) of the derivative? (c) When is the object moving to the right and when is the object moving to the left? To answer this part all we need to know is where the derivative is positive (and hence the object is moving to the right) or negative (and hence the object is moving to the left). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals.


Calculus I

Here is the number line for this problem.

From this we get the following right/left movement information.

Moving Right : 0 3, 7Moving Left : 0, 3 7

t tt t

< < < < ∞−∞ < < < <

Note that depending upon your interpretation of t as time you may or may not have included the interval 0t−∞ < < in the “Moving Left” portion. 18. Determine where the function ( ) 3 4 56 40 5 4h z z z z= + − − is increasing and decreasing.

Solution Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to get the problem started. Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. Since we are talking about where the function is increasing and decreasing we are clearly talking about the rate of change of the function. So, we’ll need the derivative.

( ) ( )( )2 3 4 2120 20 20 20 3 2h z z z z z z z′ = − − = − + − Note that the derivative was factored for later steps and doesn’t really need to be done in general. Hint : Where is the function not changing? Step 2


Calculus I

Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. ( ) ( )( )20 20 3 2 0h z z z z′ = ⇒ − + − = From this it is pretty easy to see that the derivative will be zero, and hence the function will not be moving, at,

0 3 2z z z= = − = Hint : How does the increasing/decreasing behavior of the function relate to the sign (positive or negative) of the derivative? Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.

From this we get the following increasing/decreasing information.

Increasing : 3 0, 0 2Decreasing : 3, 2

z zz z

− < < < <−∞ < < − < < ∞

19. Determine where the function ( ) ( )( )21 2R x x x= + − is increasing and decreasing.

Solution Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to get the problem started.


Calculus I

Step 1 Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. Since we are talking about where the function is increasing and decreasing we are clearly talking about the rate of change of the function. So, we’ll need the derivative. First however we’ll need to multiply out the function so we can actually take the derivative. Here is the rewritten function and the derivative.

( ) ( ) ( )3 2 23 4 3 6 3 2R x x x R x x x x x′= − + = − = − Note that the derivative was factored for later steps and doesn’t really need to be done in general. Hint : Where is the function not changing? Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. ( ) ( )0 3 2 0R x x x′ = ⇒ − = From this it is pretty easy to see that the derivative will be zero, and hence the function will not be moving, at,

0 2x x= = Hint : How does the increasing/decreasing behavior of the function relate to the sign (positive or negative) of the derivative? Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.


Calculus I


Increasing : 0, 2Decreasing : 0 2

x zz

−∞ < < < < ∞< <

20. Determine where, if anywhere, the tangent line to ( ) 3 25f x x x x= − + is parallel to the line

4 23y x= + . Solution Step 1 The first thing that we’ll need of course is the slope of the tangent line. So, all we need to do is take the derivative of the function.

( ) 23 10 1f x x x′ = − + Hint : What is the relationship between the slope of two parallel lines? Step 2 Two lines that are parallel will have the same slope and so all we need to do is determine where the slope of the tangent line will be 4, the slope of the given line. In other words, we’ll need to solve, ( ) 2 24 3 10 1 4 3 10 3 0f x x x x x′ = → − + = → − − = This quadratic doesn’t factor and so a quick use of the quadratic formula will solve this for us.

10 136 10 2 34 5 346 6 3

x ± ± ±= = =

So, the tangent line will be parallel to 4 23y x= + at,

5 34 5 340.276984 3.610323 3

x x− += = − = =


Calculus I

Product and Quotient Rule

1. Use the Product Rule to find the derivative of ( ) ( )( )2 3 24 8 12f t t t t t= − − + .

Solution There isn’t much to do here other than take the derivative using the product rule.

( ) ( )( ) ( )( )3 2 2 2 4 3 28 1 8 12 4 3 16 20 132 24 96 12f t t t t t t t t t t t t′ = − − + + − − = − + + − Note that we multiplied everything out to get a “simpler” answer.

2. Use the Product Rule to find the derivative of ( )( )3 3 31 2y x x x−= + − .

Solution There isn’t much to do here other than take the derivative using the product rule. We’ll also need to convert the roots to fractional exponents.

13

3 321 2y x x x− = + −

The derivative is then,

1 2 2 51 3 5

3 4 43 3 3 62 2 23 2 3 2 112 1 3 32 3 2 3 3

dy x x x x x x x x x xdx

− −−− − − = − + + − − = − − − −

Note that we multiplied everything out to get a “simpler” answer.

3. Use the Product Rule to find the derivative of ( ) ( )( )2 2 31 2 3 5 8h z z z z z z= + + + − .

Solution There isn’t much to do here other than take the derivative using the product rule.


Calculus I

( ) ( )( ) ( )( )2 3 2 2

2 3 4

2 6 5 8 1 2 3 5 16 3

5 36 90 88 15

h z z z z z z z z z

z z z z

′ = + + − + + + + −

= + + + −

Note that we multiplied everything out to get a “simpler” answer.

4. Use the Quotient Rule to find the derivative of ( )26

2xg x

x=

− .

Solution There isn’t much to do here other than take the derivative using the quotient rule.

( ) ( ) ( )( ) ( )

2 2

2 2

12 2 6 1 24 62 2

x x x x xg xx x

− − − −′ = =− −

5. Use the Quotient Rule to find the derivative of ( )4

2

32 1w wR ww+

=+

.


( ) ( )( ) ( )( )( ) ( )

3 2 4 5 3 2

2 22 2

3 4 2 1 3 4 4 4 6 3

2 1 2 1

w w w w w w w wR ww w

+ + − + + − +′ = =+ +

6. Use the Quotient Rule to find the derivative of ( ) 2

27 4

x xf xx x+

=−

.



Calculus I

( )( )( ) ( )( )

( )

1 121 2 22

22

7 4 7 82 27 4

x x xx x xf xx x

− − − −+ +′ =−

7. If ( )2 8f = − , ( )2 3f ′ = , ( )2 17g = and ( )2 4g′ = − determine the value of ( ) ( )2f g ′ .

Solution We know that the product rule is,

( ) ( ) ( ) ( ) ( ) ( )f g x f x g x f x g x′ ′ ′= + Now, we want to know the value of this at 2x = and so all we need to do is plug this into the derivative. Doing this gives,

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2f g f g f g′ ′ ′= + Now, we were given values for all these quantities and so all we need to do is plug these into our “formula” above.

( ) ( ) ( )( ) ( )( )2 3 17 8 4 83f g ′ = + − − = 8. If ( ) ( )3f x x g x= , ( )7 2g − = , ( )7 9g′ − = − determine the value of ( )7f ′ − .

Hint : Even though we don’t know what ( )g x is we can still use the product rule to take the

derivative and then we can use the given information to get the value of ( )7f ′ − .

Solution Even though we don’t know what ( )g x is we do have a product of two functions here and so we

can use the product rule to determine the derivative of ( )f x .

( ) ( ) ( )2 33f x x g x x g x′ ′= + Now all we need to do is plug 7x = − into this and use the given information to determine the value of ( )7f ′ − .


Calculus I

( ) ( ) ( ) ( ) ( ) ( )( ) ( )( )2 37 3 7 7 7 7 3 49 2 343 9 3381f g g′ ′− = − − + − − = + − − =

9. Find the equation of the tangent line to ( ) ( )( )21 12 4f x x x= + − at 9x = .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. We’ll use the product rule to get the derivative.

( ) ( ) ( )( ) ( ) ( )1

2 22 66 4 1 12 2 4 2 1 12f x x x x x x x xx

− ′ = − + + − = − − +

Step 2 Note that we didn’t bother to “simplify” the derivative (other than converting the fractional exponent back to a root) because all we really need this for is a quick evaluation. Speaking of which here are the evaluations that we’ll need for this problem. ( ) ( )( ) ( ) ( )( ) ( )9 37 77 2849 9 2 77 18 37 820f f ′= − = − = − − = − Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )9 9 9 2849 820 9 820 4531y f f x x y x′= + − = − − − → = − +

10. Determine where ( )2

21 8x xf x

x−

=+

is increasing and decreasing.

Solution Step 1 We’ll first need the derivative, which will require the quotient rule, because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( )( )( ) ( )( )

( ) ( )

2 2 2

2 22 2

1 2 1 8 16 1 2 8

1 8 1 8

x x x x x x xf xx x

− + − − − −′ = =+ +


Calculus I

Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. In this case it is clear that the denominator will never be zero for any real number and so the derivative will only be zero where the numerator is zero. Therefore, setting the numerator equal to zero and solving gives,

( ) ( )( )2 21 2 8 8 2 1 4 1 2 1 0x x x x x x− − = − + − = − − + = From this it is pretty easy to see that the derivative will be zero, and hence the function will not be changing, at,

1 12 4

x x= − =

Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.


1 1Increasing :2 4

1 1Decreasing : ,2 4

x

x x

− < <

−∞ < < − < < ∞

11. Determine where ( ) ( )( )2 24 1 5V t t t= − + is increasing and decreasing.


Calculus I

Solution Step 1 We’ll first need the derivative, for which we will use the product rule, because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( )( ) ( )( ) ( )2 2 3 22 1 5 4 10 38 20 2 19 10V t t t t t t t t t′ = − + + − = − = −

Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. From the factored form of the derivative it is easy to see that the derivative will be zero at,

190 1.378410

t t= = ± = ±



19 19Increasing : , 010 10

19 19Decreasing : 0,10 10

t t

t x

−∞ < < − < <

− < < < < ∞


Calculus I

Derivatives of Trig Functions

1. Evaluate ( )

0

sin 10limz

zz→

.

Solution All we need to do is set this up to allow us to use the fact from the notes in this section.

( ) ( ) ( ) ( )0 0 0

sin 10 10sin 10 sin 10lim lim 10lim 10 1 10

10 10z z z

z z zz z z→ → →

= = = =

2. Evaluate ( )( )0

sin 12lim

sin 5α

αα→

.


( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

0 0 0

0 0 0

sin 12 12 sin 12 sin 125 12 5lim lim limsin 5 12 5 sin 5 5 12 sin 5

sin 12 sin 1212 5 12 5lim lim lim5 12 sin 5 5 12 sin 5

12 121 15 5

α α α

α α α

α α α αα α αα α α α α α α

α αα αα α α α

→ → →

→ → →

= =

= =

= =

3. Evaluate ( )

0

cos 4 1limx

xx→

− .



Calculus I

( ) ( )( ) ( ) ( )0 0 0

4 cos 4 1cos 4 1 cos 4 1lim lim 4lim 4 0 0

4 4x x x

xx xx x x→ → →

−− −= = = =

4. Differentiate ( ) ( ) ( )2cos 6sec 3f x x x= − + .

Solution Not much to do here other than take the derivative. ( ) ( ) ( ) ( )2sin 6sec tanf x x x x′ = − − 5. Differentiate ( ) ( ) ( )10 tan 2cotg z z z= − .

Solution Not much to do here other than take the derivative. ( ) ( ) ( )2 210sec 2cscg z z z′ = + 6. Differentiate ( ) ( ) ( )tan secf w w w= .

Solution Not much to do here other than take the derivative, which will require the product rule.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 2sec sec tan sec tan sec sec tanf w w w w w w w w w′ = + = + 7. Differentiate ( ) ( )3 2 sinh t t t t= − .

Solution


Calculus I

Not much to do here other than take the derivative, which will require the product rule for the second term. You’ll need to be careful with the minus sign on the second term. You can either use a set of parenthesis around the derivative of the second term or you can think of the minus sign as part of the “first” function. We’ll think of the minus sign as part of the first function for this problem.

( ) ( ) ( )2 23 2 sin cosh t t t t t t′ = − −

8. Differentiate ( )6 4 cscy x x= + .

Solution Not much to do here other than take the derivative, which will require the product rule for the second term.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )1 1

1 2 224 csc 4 csc cot 2 csc 4 csc coty x x x x x x x x x x

− −′ = + − = −

9. Differentiate ( ) ( ) ( )1

2sin 4cosR t

t t=

− .

Solution Not much to do here other than take the derivative, which will require the quotient rule.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( )

( ) ( )( ) ( )( )2 2

0 2sin 4cos 1 2cos 4sin 2cos 4sin

2sin 4cos 2sin 4cos

t t t t t tR t

t t t t

− − + − −′ = =

− −

10. Differentiate ( ) ( )( )

tan1 cscv v

Z vv

+=

+ .

Solution Not much to do here other than take the derivative, which will require the quotient rule.


Calculus I

( )( )( ) ( )( ) ( )( ) ( ) ( )( )

( )( )( )( ) ( )( ) ( ) ( ) ( )( )

( )( )

2

2

2

2

1 sec 1 csc tan csc cot

1 csc

1 sec 1 csc csc cot tan

1 csc

v v v v v vZ v

v

v v v v v v

v

+ + − + −′ =

+

+ + + +=

+

11. Find the tangent line to ( ) ( ) ( )tan 9cosf x x x= + at x π= .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. ( ) ( ) ( )2sec 9sinf x x x′ = − Step 2 Now all we need to do is evaluate the function and the derivative at the point in question. ( ) ( ) ( ) ( ) ( ) ( )2tan 9cos 9 sec 9sin 1f fπ π π π π π′= + = − = − = Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )( )9 1 9y f f x x y xπ π π π π′= + − = − + − → = − − Don’t get excited about the presence of the π in the answer. It is just a number like the 9 is and so is nothing to worry about. 12. The position of an object is given by ( ) ( )2 7coss t t= + determine all the points where the

object is not moving. Solution We know that the object will not be moving if its velocity, which is simply the derivative of the position function, is not zero. So all we need to do is take the derivative, set it equal to zero and solve.


Calculus I

( ) ( ) ( )7sin 7sin 0s t t t′ = − ⇒ − = So, for this problem the object will not be moving anywhere that sine is zero. From our recollection of the unit circle we know that will be at,

0 2 2 and 2 0, 1, 2, 3,t n n t n nπ π π π= + = = + = ± ± ± 13. Where in the range [ ]2,7− is the function ( ) ( )4cosf x x x= − is increasing and

decreasing. Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( )4sin 1f x x′ = − − Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve.

( ) ( ) 14sin 1 0 sin4

x x− − = ⇒ = −

A quick calculator computation tells us that,

1 1sin 0.25274

x − = − = −

Recalling our work in the Review chapter on solving trig equations we know that a positive angle corresponding to this solution is : 2 0.2527 6.0305x π= − = . Either can be used, but we will use the positive angle. Also, from a quick check on a unit circle we can see that 0.2527 3.3943x π= + = will be a second solution. Putting all of this together and we can see that the derivative will be zero at, 6.0305 2 and 3.3943 2 0, 1, 2, 3,x n x n nπ π= + = + = ± ± ±


Calculus I

Finally, all we need to so is plug in some n’s to determine which solutions fall in the interval we are working on, [ ]2,7− .

1: 0.2527 2.8889n x x= − = − = −

0 : 6.0305 3.39431: 12.3137

n x xn x= = == = 9.6775x =

So, in the interval [ ]2,7− the function will stop changing at the following three points.

0.2527, 3.3943, 6.0305x = − Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.


Increasing : 2 0.2527, 3.3943 6.0305Decreasing : 0.2527 3.3943, 6.0305 7

x xx x

− ≤ < − < <− < < < ≤

Note that because we’ve only looked at what is happening in the interval [ ]2,7− we can’t say

anything about the increasing/decreasing nature of the function outside of this interval.


Calculus I

Derivatives of Exponential and Logarithm Functions 1. Differentiate ( ) 2 8xxf x = −e .

Solution Not much to do here other than take the derivative using the formulas from class.

( ) ( )2 8 ln 8xxf x′ = −e 2. Differentiate ( ) ( ) ( )34 log lng t t t= − .


( ) ( )4 1

ln 3g t

t t′ = −

3. Differentiate ( ) ( )3 logwR w w= .


( ) ( ) ( ) ( )33 ln 3 log

ln 10

wwR w w

w′ = +

Recall that ( )log x is the common logarithm and so is really ( )10log x .

4. Differentiate ( )5 lnzy z z= − e .


Calculus I


( )45 lnz

zy z zz

′ = − −ee

5. Differentiate ( )1 y

yh y =− e

.


( )( )( ) ( )

( ) ( )2 2

1 1 1

1 1

y y y y

y y

y yh y− − − − +′ = =

− −

e e e e

e e

6. Differentiate ( ) ( )1 5ln

tf tt

+= .


( ) ( )

( ) ( )

( )

( )

( )2 2

1 15ln 1 5 5ln 51 5ln ln ln

t t tt t tf tt t t

− + − − + = = =

7. Find the tangent line to ( ) 7 4x xf x = + e at 0x = .

Solution Step 1


Calculus I

We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. ( ) ( )7 ln 7 4x xf x′ = + e Step 2 Now all we need to do is evaluate the function and the derivative at the point in question. ( ) ( ) ( )0 5 0 ln 7 4 5.9459f f ′= = + = Step 3 Now all that we need to do is write down the equation of the tangent line.

( ) ( )( ) ( )( )0 0 0 5 ln 7 4 5 5.9459y f f x x x′= + − = + + = +

8. Find the tangent line to ( ) ( ) ( )2ln logf x x x= at 2x = .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function.

( ) ( ) ( )( )

2log lnln 2

x xf x

x x′ = +

Step 2 Now all we need to do is evaluate the function and the derivative at the point in question.

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

22

log 2 ln 22 ln 2 log 2 ln 2 2 1

2 2ln 2f f ′= = = + =


( ) ( )( ) ( ) ( )( ) ( )2 2 2 ln 2 1 2 2 ln 2y f f x x x′= + − = + − = − +

9. Determine if ( ) ttV t =e

is increasing or decreasing at the following points.

(a) 4t = − (b) 0t = (c) 10t = Solution


Calculus I

(a) 4t = − We know that the derivative of the function will give us the rate of change for the function and so we’ll need that.

( )( ) ( )

( ) ( )2 2

1 1t t t t

tt t

t t tV t− − −′ = = =

e e e eee e

Now, all we need to do is evaluate the derivative at the point in question. So,

( ) 454 272.991 0V −′ − = = >

e

( )4 0V ′ − > and so the function must be increasing at 4t = − .

(b) 0t = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) 010 1 0V ′ = = >e

( )0 0V ′ > and so the function must be increasing at 0t = .

(c) 10t = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) 10910 0.0004086 0V −′ = = − <

e

( )10 0V ′ < and so the function must be decreasing at 10t = .

10. Determine if ( ) ( ) ( )6 lnG z z z= − is increasing or decreasing at the following points.

(a) 1z = (b) 5z = (c) 20z = Solution (a) 1z = We know that the derivative of the function will give us the rate of change for the function and so we’ll need that.


Calculus I

( ) ( ) 6ln zG z zz−′ = +

Now, all we need to do is evaluate the derivative at the point in question. So, ( ) ( )1 ln 1 5 5 0G′ = − = − <

( )1 0G′ < and so the function must be decreasing at 1z = .

(b) 5z = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) ( ) 15 ln 5 1.40944 05

G′ = − = >

( )5 0G′ > and so the function must be increasing at 5z = .

(c) 20z = We found the derivative of the function in the first part so here all we need to do is the evaluation.

( ) ( ) 720 ln 20 3.6957310

G′ = + =

( )20 0G′ > and so the function must be increasing at 20z = .

Derivatives of Inverse Trig Functions 1. Differentiate ( ) ( ) ( )12cos 6cosT z z z−= + .


( ) ( )2

62sin1

T z zz

′ = − −−


Calculus I

2. Differentiate ( ) ( ) ( )1 1csc 4cotg t t t− −= − .


( ) 22

1 411

g ttt t

′ = − ++−

3. Differentiate ( )6 15 secy x x−= − .


5

2

1301

dy xdx x x

= −−

4. Differentiate ( ) ( ) ( )2 1sin tanf w w w w−= + .


( ) ( ) ( )2

12cos 2 tan

1wf w w w w

w−′ = + +

+

5. Differentiate ( ) ( )1sin1

xh x

x

−

=+

.



Calculus I

( )( )

( )( )

( )

12 12

2 22

1 sin1 1 sin1

1 1 1

x xx x xxh x

x x x

−−

+−

+ − −−′ = =+ − +

Derivatives of Hyperbolic Functions 1. Differentiate ( ) ( ) ( ) ( )sinh 2cosh sechf x x x x= + − .


( ) ( ) ( ) ( ) ( )cosh 2sinh sech tanhf x x x x x′ = + + 2. Differentiate ( ) ( ) ( )2tan cschR t t t t= + .


( ) ( ) ( ) ( ) ( )2 2sec 2 csch csch cothR t t t t t t t′ = + −

3. Differentiate ( ) ( )1

tanhzg z

z+

= .



Calculus I

( ) ( ) ( ) ( )( )

2

2

tanh 1 sechtanh

z z zg z

z− +

′ =

Chain Rule

1. Differentiate ( ) ( )426 7f x x x= + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. The derivative is then,

( ) ( ) ( ) ( )( )3 32 24 6 7 12 7 4 12 7 6 7f x x x x x x x′ = + + = + +

2. Differentiate ( ) ( ) 224 3 2g t t t−

= − + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the exponent of -2 on the parenthesis while the inside function is the polynomial that is being raised to the power. The derivative is then,

( ) ( ) ( ) ( )( )3 32 22 4 3 2 8 3 2 8 3 4 3 2g t t t t t t t− −

′ = − − + − = − − − +


Calculus I

3. Differentiate 3 1 8y z= − . Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem, after converting the root to a fractional exponent, the outside function is

(hopefully) clearly the exponent of 13

while the inside function is the polynomial that is being

raised to the power (or the polynomial inside the root – depending upon how you want to think about it). The derivative is then,

( ) ( ) ( ) ( )1 2 23 3 3

1 81 8 1 8 8 1 83 3

dyy z z zdz

− −= − ⇒ = − − = − −

4. Differentiate ( ) ( )csc 7R w w= .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the trig function and the inside function is the stuff inside of the trig function. The derivative is then,

( ) ( ) ( )7csc 7 cot 7R w w w′ = − In dealing with functions like cosecant (or secant for that matter) be careful to make sure that the inside function gets substituted into both terms of the derivative of the outside function. One of the more common mistakes with this kind of problem is to only substitute the 7w into only the cosecant or only the cotangent instead of both as it should be. 5. Differentiate ( ) ( )( )2sin 3 tanG x x x= + .


Calculus I

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the sine function and the inside function is the stuff inside of the trig function. The derivative is then,

( ) ( )( ) ( )( )22 3 sec cos 3 tanG x x x x′ = + +

6. Differentiate ( ) ( )tan 4 10h u u= + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the trig function and the inside function is the stuff inside of the trig function. The derivative is then,

( ) ( )210sec 4 10h u u′ = +

7. Differentiate ( ) 745 t tf t += + e .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution Note that we only need to use the Chain Rule on the second term as we can differentiate the first term without the Chain Rule. Now, recall that for exponential functions outside function is the exponential function itself and the inside function is the exponent. The derivative is then,

( ) ( ) 76 44 7 t tf t t +′ = + e


Calculus I

8. Differentiate ( ) ( )1 cos xg x −= e .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For exponential functions remember that the outside function is the exponential function itself and the inside function is the exponent. The derivative is then,

( ) ( ) ( )1 cossin xg x x −′ = e 9. Differentiate ( ) 1 62 zH z −= .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For exponential functions remember that the outside function is the exponential function itself and the inside function is the exponent. The derivative is then,

( ) ( ) ( )1 66 2 ln 2zH z −′ = −

10. Differentiate ( ) ( )1tan 3 1u t t−= − .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the inverse tangent and the inside function is the stuff inside of the inverse tangent. The derivative is then,


Calculus I

( )( )2

33 1 1

u tt

′ =− +

11. Differentiate ( ) ( )2 3ln 1 5F y y y= − + .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the logarithm and the inside function is the stuff inside of the logarithm. The derivative is then,

( ) ( )2

22 3 2 3

1 10 310 31 5 1 5

y yF y y yy y y y

− += − + =

− + − +

With logarithm problems remember that after differentiating the logarithm (i.e. the outside function) you need to substitute the inside function into the derivative. So, instead of getting just,

1y

we get the following (i.e. we plugged the inside function into the derivative),

2 3

11 5y y− +

Then, we can’t forget of course to multiply by the derivative of the inside function. 12. Differentiate ( ) ( ) ( )( )ln sin cotV x x x= − .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Solution For this problem the outside function is (hopefully) clearly the logarithm and the inside function is the stuff inside of the logarithm. The derivative is then,


Calculus I

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )

22 cos csc1 cos csc

sin cot sin cotx x

V x x xx x x x

+= + =

− −

With logarithm problems remember that after differentiating the logarithm (i.e. the outside function) you need to substitute the inside function into the derivative. So, instead of getting just,

1x

we get the following (i.e. we plugged the inside function into the derivative),

( ) ( )

1sin cotx x−

Then, we can’t forget of course to multiply by the derivative of the inside function.

13. Differentiate ( ) ( ) ( )6 6sin sinh z z z= + .

Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem each term will require a separate application of the Chain Rule and don’t forget that,

( ) ( ) 66sin sinz z= So, in the first term the outside function is the sine function, while the sine function is the inside function in the second term. The derivative is then,

( ) ( ) ( ) ( )5 6 56 cos 6sin cosh z z z z z′ = +

14. Differentiate ( ) 7 wS w w −= + e .


Calculus I

Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem each term will require a separate application of the Chain Rule and make sure you are careful with parenthesis in dealing with the root in the first term. The derivative is then,

( ) ( ) ( ) ( )( ) ( )1 1 12 2 2

1 77 7 7 72 2

w w wS w w S w w w− −− − −′= + ⇒ = − = −e e e

15. Differentiate ( ) ( )7 23 sin 6g z z z= − + .

Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem the first term requires no Chain Rule and the second term will require the Chain Rule. The derivative is then,

( ) ( )6 221 2 cos 6g z z z z′ = − +

16. Differentiate ( ) ( )( ) ( )104ln sin 3f x x x x= − − .

Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule. Sometimes separate terms will require different applications of the Chain Rule, or maybe only one of the terms will require the Chain Rule. Solution For this problem each term will require a separate application of the Chain Rule. The derivative is then,


Calculus I

( ) ( )( ) ( )( ) ( ) ( )( )9 93 4 3 4cos

10 4 3 3 cot 10 4 3 3sin

xf x x x x x x x x

x′ = − − − = − − −

17. Differentiate ( ) 6 25h t t t t= − .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate the second term. The derivative is then,

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( )

16 2 2

1 1 1 15 2 6 2 5 2 6 22 2 2 2

5

1 16 5 5 10 1 6 5 10 1 52 2

h t t t t

h t t t t t t t t t t t t t t t− −

= −

′ = − + − − = − + − −

18. Differentiate ( ) ( )2 5lnq t t t= .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate the second term. The derivative is then,


Calculus I

( ) ( ) ( )4

5 2 55

52 ln 2 ln 5tq t t t t t t tt

′ = + = +

19. Differentiate ( ) ( ) ( )cos 3 sec 1g w w w= − .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate each term. The derivative is then,

( ) ( )( ) ( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )sin 3 3 sec 1 cos 3 sec 1 tan 1 1

3sin 3 sec 1 cos 3 sec 1 tan 1

g w w w w w w

w w w w w

′ = − − + − − −

= − − − − −

20. Differentiate ( )

2

sin 31

ty

t=

+ .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Quotient Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate the numerator. The derivative is then,

( )( ) ( )( )

( )( )( ) ( )

( )

2 2

2 22 2

3cos 3 1 sin 3 2 3cos 3 1 2 sin 3

1 1

t t t t t t t tdydt t t

+ − + −= =

+ +


Calculus I

21. Differentiate ( ) ( )21

tan 12

xK x

x x

−+=

+e

.

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient. Solution For this problem we’ll need to do the Quotient Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate both the numerator and the denominator. The derivative is then,

( )( )( ) ( ) ( )( )

( )( )

2 2 2

2

2 tan 12 1 1 12sec 12

tan 12

x xx x xK x

x x

− −− + − + +′ =

+

e e

22. Differentiate ( ) ( )2cos xf x x= e .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of using the Chain Rule, you’ll also need the Product and/or Quotient Rule. Solution For this problem we’ll start off using the Chain Rule, however when we differentiate the inside function we’ll need to do the Product Rule. The derivative is then,

( ) ( ) ( )2 22 sinx x xf x x x x′ = − +e e e

23. Differentiate ( )5 tan 4z x x= + .


Calculus I

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.

( )( )

( )( ) ( )( )

12

12

5 tan 4

1 5 tan 4 5 tan 42

z x x

dz dx x x xdx dx

−

= +

= + +

Step 2 In this step we can see that we’ll need to use the Chain Rule on the second term. The derivative is then,

( )( ) ( )( )1

221 5 tan 4 5 4sec 42

dz x x xdx

−= + +

In this step we were using the Chain Rule on the second term and so when multiplying by the derivative of the inside function we only multiply the second term by the derivative of the inside function and not both terms.

24. Differentiate ( ) ( )( )36 sin 2tf t t−= + −e .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear.


Calculus I

Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.

( ) ( )( ) ( )( )26 63 sin 2 sin 2t tdf t t tdt

− −′ = + − + −e e Step 2 In this step we can see that we’ll need to use the Chain Rule on each of the terms. The derivative is then,

( ) ( )( ) ( )( )26 63 sin 2 6 cos 2t tf t t t− −′ = + − − − −e e

25. Differentiate ( ) ( ) ( )( )102 1ln 1 tan 6g x x x−= + − .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.

( ) ( ) ( )( ) ( ) ( )( )92 1 2 110 ln 1 tan 6 ln 1 tan 6dg x x x x xdx

− −′ = + − + − Step 2 In this step we can see that we’ll need to use the Chain Rule on each of the terms. The derivative is then,

( ) ( ) ( )( )92 12 2

2 610 ln 1 tan 61 36 1

xg x x xx x

− ′ = + − − + +


Calculus I

26. Differentiate ( ) ( )4 2tan 1h z z= + .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. Also, recall that,

( ) ( ) 44tan tanx x=


( ) ( ) ( )3 2 24 tan 1 tan 1dh z z zdz

′ = + + Step 2 As we can see the derivative from the previous step will also require the Chain Rule. The derivative is then,

( ) ( ) ( )( ) ( ) ( )3 2 2 2 3 2 2 24 tan 1 sec 1 2 8 tan 1 sec 1h z z z z z z z′ = + + = + +

27. Differentiate ( ) ( )( ) 123 12 sin 3f x x x

−= + .

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative. Step 1 This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear.


Calculus I


( ) ( )( ) ( ) ( )12

2 23 312 sin 3 12 sin 3df x x x x xdx

− ′ = − + +

Step 2 As we can see the derivative from the previous step will also require the Chain Rule on each of the terms. The derivative from this step is,

( ) ( )( ) ( ) ( ) ( ) ( )( )22

23 3112 sin 3 12 12 2sin 3 sin 33

df x x x x x xdx

− − ′ = − + +

Step 3 The second term will again use the Chain Rule as we can see. The derivative is then,

( ) ( )( ) ( ) ( ) ( )22

23 312 sin 3 4 12 6sin 3 cos 3f x x x x x x− − ′ = − + +

28. Find the tangent line to ( ) 24 2 6 xf x x −= − e at 2x = .

Solution Step 1 We know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. Differentiating each term will require the Chain Rule as well.

( ) ( )

( ) ( ) ( ) ( ) ( )

122

1 12 2 22 2

4 2 6

1 44 2 2 6 1 4 2 6 62 2

x

x x x

f x x

f x x xx

−

− −− − −

= −

′ = − − = + = +

e

e e e

Step 2 Now all we need to do is evaluate the function and the derivative at the point in question.

( ) ( ) ( )0 042 4 2 6 2 2 6 82

f f ′= − = = + =e e


Calculus I


( ) ( )( ) ( )2 2 2 2 8 2 8 14y f f x x y x′= + − = + − → = −

29. Determine where ( ) ( )34 2 8V z z z= − is increasing and decreasing.

Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( )

3 23 4

2 23 3

4 2 8 3 2 8 2

2 2 8 2 2 8 3 2 2 8 7 16

V z z z z z

z z z z z z z

′ = − + −

= − − + = − −

Note that we factored the derivative to help with the next step. In general we don’t need to do this. Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. In this case it’s pretty easy to spot where the derivative will be zero.

( ) ( )23 162 2 8 7 16 0 0, 4, 2.28577

z z z z z z− − = ⇒ = = = =



Calculus I


16Increasing : 0, 4, 47

16Decreasing : 07

x x x

x

−∞ < < < < < < ∞

< <

30. The position of an object is given by ( ) ( )sin 3 2 4s t t t= − + . Determine where in the

interval [ ]0,3 the object is moving to the right and moving to the left.

Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the velocity of the object. Here is the derivative.

( ) ( )3cos 3 2s t t′ = − Step 2 Next, we need to know where the object stops moving and so all we need to do is set the derivative equal to zero and solve.

( ) ( ) 23cos 3 2 0 cos 33

t t− = ⇒ =

A quick calculator computation tells us that,

1 23 cos 0.84113

t − = =


Calculus I

Recalling our work in the Review chapter and a quick check on a unit circle we can see that either 3 0.8411t = − or 3 2 0.8411 5.4421t π= − = can be used for the second angle. Note that either will work, but we’ll use the second simply because it is the positive angle. Putting all of this together and dividing by 3 we can see that the derivative will be zero at,

3 0.8411 2 and 3 5.4421 2 0, 1, 2, 3,

2 20.2804 and 1.8140 0, 1, 2, 3,3 3

t n t n nn nt t n

π ππ π

= + = + = ± ± ±

= + = + = ± ± ±

Finally, all we need to so is plug in some n’s to determine which solutions fall in the interval we are working on, [ ]0,6 .

0 : 0.2804 1.81401: 2.3748 3.9084

n t tn t t= = == = =

So, in the interval [ ]0,6 , the object stops moving at the following three points.

0.2804, 1.8140, 2.3748t = Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the object is moving to the right) or negative (and hence the object is moving to the left). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.

From this we get the following moving right/moving left information.

Moving Right : 0 0.2804, 1.8140 2.3748Moving Left : 0.2804 1.8140, 2.3748 3

t xx x

≤ < < << < < ≤


Calculus I

Note that because we’ve only looked at what is happening in the interval [ ]0,3 we can’t say

anything about the moving right/moving left behavior of the object outside of this interval. 31. Determine where ( ) 2 5 tA t t −= e is increasing and decreasing.

Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( )5 2 5 52 2t t tA t t t t t− − −′ = − = −e e e Note that we factored the derivative to help with the next step. In general we don’t need to do this. Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve. In this case it’s pretty easy to spot where the derivative will be zero. ( )5 2 0 0, 2tt t t t− − = ⇒ = =e Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem.



Calculus I

Increasing :0 2Decreasing : 0, 2

tt t

< <−∞ < < < < ∞

32. Determine where in the interval [ ]1,20− the function ( ) ( )4 3ln 20 100f x x x= − − is

increasing and decreasing. Solution Step 1 We’ll first need the derivative because we know that the derivative will give us the rate of change of the function. Here is the derivative.

( ) ( )23 2

4 3 4 3

4 154 6020 100 20 100

x xx xf xx x x x

−−= =

− − − −

Note that we factored the numerator to help with the next step. In general we don’t need to do this. Step 2 Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve.

( ) ( )

22

4 3

4 150 4 15 0 0, 15

20 100x x

x x x xx x

−= → − = ⇒ = =

− −

Note, that in general, we also need to be concerned with where the derivative is not defined as well. Functions can (and often do) change sign where they are not defined. In this case however we’ve restricted the interval down to a range where the function exists and is continuous on the given interval and so this is something we need to worry about for this problem. In the next Chapter we will start also looking at what happens if the derivative is also not defined at particular points. Note as well that we really should at this point step back and recall that we are working on the interval [ ]1,20− . We are only interested in what is happening on this interval and so we should

make sure that the points found above are inside the interval. In this case both points are inside the interval we are interested in. However, if any of them had been outside of the interval we would have needed to exclude that point.


Calculus I



Increasing : 1 0, 0 15Decreasing : 15 20

x xx

− ≤ < < << ≤

Note that because we’ve only looked at what is happening in the interval [ ]1,20− we can’t say

anything about the increasing/decreasing nature of the function outside of this interval.

Implicit Differentiation

1. For 3 1xy

= do each of the following.

(a) Find y′ by solving the equation for y and differentiating directly. (b) Find y′ by implicit differentiation. (c) Check that the derivatives in (a) and (b) are the same.

(a) Find y′ by solving the equation for y and differentiating directly. Step 1 First we just need to solve the equation for y.


Calculus I

1

3 3y x y x= ⇒ = Step 2 Now differentiate with respect to x.

2

1 33y x

−′ =

(b) Find y′ by implicit differentiation.

Hint : Don’t forget that y is really ( )y x and so we’ll need to use the Chain Rule when taking the

derivative of terms involving y! Also, don’t forget that because y is really ( )y x we may well

have a Product and/or a Quotient Rule buried in the problem. Step 1 First, we just need to take the derivative of everything with respect to x and we’ll need to recall that y is really ( )y x and so we’ll need to use the Chain Rule when taking the derivative of terms

involving y. Also, prior to taking the derivative a little rewrite might make this a little easier. 3 1x y− = Now take the derivative and don’t forget that we actually have a product of functions of x here and so we’ll need to use the Product Rule when differentiating the left side. 3 43 0y x y y− − ′− = Step 2 Finally all we need to do is solve this for y′ .

3

43 3y yyxy x

−

−′ = =

(c) Check that the derivatives in (a) and (b) are the same. Hint : To show they are the same all we need is to plug the formula for y (which we already have….) into the derivative we found in (b) and, potentially with a little work, show that we get the same derivative as we got in (a). From (a) we have a formula for y written explicitly as a function of x so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a).

123

1 333 3

y xy xx x

−′ = = =


Calculus I

So, we got the same derivative as we should. 2. For 2 3 4x y+ = do each of the following.

(d) Find y′ by solving the equation for y and differentiating directly. (e) Find y′ by implicit differentiation. (f) Check that the derivatives in (a) and (b) are the same.


( )1

3 2 2 34 4y x y x= − ⇒ = − Step 2 Now differentiate with respect to x.

( )2

22 33 4y x x

−′ = − −

(b) Find y′ by implicit differentiation.


derivative of terms involving y! Step 1 First, we just need to take the derivative of everything with respect to x and we’ll need to recall that y is really ( )y x and so we’ll need to use the Chain Rule when taking the derivative of terms

involving y. Taking the derivative gives, 22 3 0x y y′+ = Step 2 Finally all we need to do is solve this for y′ .

2

23

xyy

′ = −

(c) Check that the derivatives in (a) and (b) are the same. Hint : To show they are the same all we need is to plug the formula for y (which we already have….) into the derivative we found in (b) and, potentially with a little work, show that we get the same derivative as we got in (a).


Calculus I

From (a) we have a formula for y written explicitly as a function of x so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a).

( )

( )2

22 3322

2 3

2 2 43 3 4

x xy x xy x

−′ = − = − = − −

−

So, we got the same derivative as we should. 3. For 2 2 2x y+ = do each of the following.

(g) Find y′ by solving the equation for y and differentiating directly. (h) Find y′ by implicit differentiation. (i) Check that the derivatives in (a) and (b) are the same.


( )1

2 2 2 22 2y x y x= − ⇒ = ± −

Note that because we have no restriction on y (i.e. we don’t know if y is positive or negative) we really do need to have the “± ” there and that does lead to issues when taking the derivative. Hint : Two formulas for y and so two derivatives. Step 2 Now, because there are two formulas for y we will also have two formulas for the derivative, one for each formula for y. The derivatives are then,

( ) ( ) ( )

( ) ( ) ( )

1 12 22 2

1 12 22 2

2 2 0

2 2 0

y x y x x y

y x y x x y

−

−

′= − ⇒ = − − >

′= − − ⇒ = − <

As noted above the first derivative will hold for 0y > while the second will hold for 0y < and we can use either for 0y = as the plus/minus won’t affect that case. (b) Find y′ by implicit differentiation.


Calculus I



involving y. Taking the derivative gives, 2 2 0x y y′+ = Step 2 Finally all we need to do is solve this for y′ .

xyy

′ = −

(c) Check that the derivatives in (a) and (b) are the same. Hint : To show they are the same all we need is to plug the formula for y (which we already have….) into the derivative we found in (b) and, potentially with a little work, show that we get the same derivative as we got in (a). Again, two formulas for y so two derivatives… From (a) we have a formula for y written explicitly as a function of x so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a). Also, because we have two formulas for y we will have two formulas for the derivative. First, if 0y > we will have,

( )( )

( )1 1

2 22 21

2 2

2 22

x xy x y x xy x

−′= − ⇒ = − = − = − −

−

Next, if 0y < we will have,

( )( )

( )1 1

2 22 21

2 2

2 22

x xy x y x xy x

−′= − − ⇒ = − = − = −

− −

So, in both cases, we got the same derivative as we should. 4. Find y′ by implicit differentiation for 3 2 62 4y x y x+ − = .


Calculus I



involving y. Differentiating with respect to x gives, 2 56 8 6y y x y x′ ′+ − = Step 2 Finally all we need to do is solve this for y′ .

( )5

2 52

6 86 1 6 86 1x xy y x x yy−′ ′− = − ⇒ =−

5. Find y′ by implicit differentiation for ( )2 47 sin 3 12y x y+ = − .



involving y. Differentiating with respect to x gives, ( ) 314 3cos 3 4y y x y y′ ′+ = − Step 2 Finally all we need to do is solve this for y′ .

( ) ( ) ( )33

3cos 314 4 3cos 3

14 4x

y y y x yy y

−′ ′+ = − ⇒ =

+

6. Find y′ by implicit differentiation for ( )sinx y x− =e .


derivative of terms involving y! Step 1


Calculus I

First, we just need to take the derivative of everything with respect to x and we’ll need to recall that y is really ( )y x and so we’ll need to use the Chain Rule when taking the derivative of terms

involving y. Differentiating with respect to x gives, ( )cos 1x y y′− =e Don’t forget the y′ on the cosine after differentiating. Again, y is really ( )y x and so when

differentiating ( )sin y we really differentiating ( )sin y x and so we are differentiating using

the Chain Rule! Step 2 Finally all we need to do is solve this for y′ .

( ) ( ) ( )1 1 sec

cos

xxy y

y−′ = = −

−e e

7. Find y′ by implicit differentiation for 2 7 5 34 2 4x y x x y− = + . Hint : Don’t forget that y is really ( )y x and so we’ll need to use the Chain Rule when taking the



involving y. This also means that the first term on the left side is really a product of functions of x and hence we will need to use the Product Rule when differentiating that term. Differentiating with respect to x gives, 7 2 6 4 28 28 2 5 12xy x y y x y y′ ′+ − = + Step 2 Finally all we need to do is solve this for y′ .

( )7 4

7 4 2 2 62 2 6

8 5 28 5 2 12 2812 28

xy xxy x y x y y yy x y− −′ ′− − = − ⇒ =−

8. Find y′ by implicit differentiation for ( ) 22cos 2 1yx y x+ + =e .


Calculus I




involving y. This also means that the second term on the left side is really a product of functions of x and hence we will need to use the Product Rule when differentiating that term. Differentiating with respect to x gives,

( ) ( ) 2 222 2 sin 2 2 0y yx y x y y y x′ ′− + + + + =e e Step 2 Finally all we need to do is solve this for y′ (with some potentially messy algebra).

( ) ( )( )( ) ( )

( )( )

2 2

2 2

2

2

2 2

2 2

2

2

2 sin 2 2 sin 2 2 0

2 2sin 2 0 2 sin 2

2 sin 2

2 2sin 2

y y

y y

y

y

x x y y x y y y x

yx x y y x x y

x x yy

yx x y

′ ′− + − + + + =

′− + = + + −

+ −′ =

− +

e e

e e

e

e

9. Find y′ by implicit differentiation for ( )2 4 2tan 3x y x y= + .




involving y. This also means that the when doing Chain Rule on the first tangent on the left side we will need to do Product Rule when differentiating the “inside term”. Differentiating with respect to x gives,

( ) ( )4 2 3 2 2 42 4 sec 3 2x y x y y x y y y′ ′+ = + Step 2 Finally all we need to do is solve this for y′ (with some potentially messy algebra).


Calculus I

( ) ( )( )( ) ( )

( )( )

4 2 2 4 2 3 2 2 4

2 3 2 2 4 4 2 2 4

4 2 2 4

2 3 2 2 4

2 sec 4 sec 3 2

4 sec 2 3 2 sec

3 2 sec

4 sec 2

x y x y x y y x y y y

x y x y y y x y x y

x y x yy

x y x y y

′ ′+ = +

′− = −

−′ =

−

10. Find the equation of then tangent line to 4 2 3x y+ = at ( )1, 2− .

Hint : We know how to compute the slope of tangent lines and with implicit differentiation that shouldn’t be too hard at this point. Step 1 The first thing to do is use implicit differentiation to find y′ for this function.

3

3 24 2 0 xx y y yy

′ ′+ = ⇒ = −

Step 2 Evaluating the derivative at the point in question to get the slope of the tangent line gives,

1, 2

2 22x y

m y= =−

′= = − =−

Step 3 Now, we just need to write down the equation of the tangent line.

( ) ( ) ( ) ( )2 2 1 2 1 2 2 2y x y x x− − = − ⇒ = − − = −

11. Find the equation of then tangent line to 2 2 23xy y x= +e at ( )0,3 .

Hint : We know how to compute the slope of tangent lines and with implicit differentiation that shouldn’t be too hard at this point. Step 1 The first thing to do is use implicit differentiation to find y′ for this function.

2 2

2 2 22

2 22 2 3 22 3

xx x

xx yyy y y x y

y−′ ′ ′+ = + ⇒ =

−ee e

e

Step 2 Evaluating the derivative at the point in question to get the slope of the tangent line gives,


Calculus I

0, 3

18 63x y

m y= =

−′= = = −

Step 3 Now, we just need to write down the equation of the tangent line.

( )3 6 0 6 3y x y x− = − − ⇒ = − + 12. Assume that ( )x x t= , ( )y y t= and ( )z z t= and differentiate 2 3 4 1x y z− + = with

respect to t. Hint : This is just implicit differentiation like we’ve been doing to this point. The only difference is that now all the functions are functions of some fourth variable, t. Outside of that there is nothing different between this and the previous problems. Solution Differentiating with respect to t gives,

2 32 3 4 0x x y y z z′ ′ ′− + =

Note that because we were not asked to give the formula for a specific derivative we don’t need to go any farther. We could however, if asked, solved this for any of the three derivatives that are present.

13. Assume that ( )x x t= , ( )y y t= and ( )z z t= and differentiate ( ) ( )2 3cos sin 4x y y z= +

with respect to t. Hint : This is just implicit differentiation like we’ve been doing to this point. The only difference is that now all the functions are functions of some fourth variable, t. Outside of that there is nothing different between this and the previous problems. Solution Differentiating with respect to t gives,

( ) ( ) ( ) ( )2 2 32 cos sin 3 4 cos 4x x y x y y y y z y z′ ′ ′ ′− = + +

Note that because we were not asked to give the formula for a specific derivative we don’t need to go any farther. We could however, if asked, solved this for any of the three derivatives that are present.


Calculus I

Related Rates 1. In the following assume that x and y are both functions of t. Given 2x = − , 1y = and 4x′ = − determine y′ for the following equation.

2 2 3 4 46 2 yy x x −+ = − e Hint : This is just like the problems worked in the section notes. The only difference is that you’ve been given the equation and all the needed information and so you won’t have to worry about finding that. Step 1 The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to t. 2 4 4 3 4 412 2 3 4y yy y x x x x x y− −′ ′ ′ ′+ = − +e e Step 2 All we need to do now is plug in the given information and solve for y′ .

81112 16 48 32y y y′ ′ ′+ = − ⇒ =

2. In the following assume that x, y and z are all functions of t. Given 4x = , 2y = − , 1z = ,

9x′ = and 3y′ = − determine z′ for the following equation.

( ) 3 2 2 21 5 3x y z y z x− + = + − Hint : This is just like the problems worked in the section notes. The only difference is that you’ve been given the equation and all the needed information and so you won’t have to worry about finding that. Step 1 The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to t. ( ) 2 2 21 15 2 2 2x y x y z z y y z y z z x x′ ′ ′ ′ ′ ′− − + = + + Step 2 All we need to do now is plug in the given information and solve for z′ .


Calculus I

45727 12 15 12 8 72z z z′ ′ ′+ + = + + ⇒ =

3. For a certain rectangle the length of one side is always three times the length of the other side.

(a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing?

(b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute?

Hint : The equation needed here is a really simple equation. In fact, so simple it might be easy to miss… (a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing? Step 1 Let’s call the shorter side x and the longer side y. We know that 2x′ = − and want to find y′ . Now all we need is an equation that relates these two quantities and from the problem statement we know the longer side is three times shorter side and so the equation is, 3y x= Step 2 Next step is to simply differentiate the equation with respect to t. 3y x′ ′=

Step 3

Finally, plug in the known quantity and solve for what we want : 6y′ = −

Hint : Once we have the equation for the area we can either simplify the equation as we did in this section or we can use the result from the previous step and the equation directly. (b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute? Step 1 Again we’ll call the shorter side x and the longer side y as with the last part. We know that 6x = ,

2x′ = − and want to find A′ . The equation we’ll need is just the area formula for a rectangle : A xy= At this point we can either leave the equation as is and differentiate it or we can plug in 3y x= to simplify the equation down to a single variable then differentiate. Doing this gives,


Calculus I

( ) 23A x x= Step 2 Now we need to differentiate with respect to t. If we use the equation in terms of only x, which is probably the easiest to use we get, 6A x x′ ′=

If we use the equation in terms of both x and y we get, A x y x y′ ′ ′= + Step 3 Now all we need to do is plug in the known quantities and solve for A′ . Using the equation in terms of only x is the “easiest” because we already have all the known quantities from the problem statement itself. Doing this gives,

( )( )6 6 2 72A′ = − = − Now let’s use the equation in terms of x and y. We know that 6x = and 2x′ = − from the

problem statement. From part (a) we have 6y′ = − and we also know that ( )3 6 18y = = .

Using these gives,

( )( ) ( )( )6 6 2 18 72A′ = − + − = − So, as we can see both gives the same result but the second method is slightly more work, although not much more. 4. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2 ? Step 1 We’ll call the area of the sheet A and the radius r and we know that the area of a circle is given by,

2A rπ= We know that 0.5A′ = − and want to determine r′ when 12A = . Step 2 Next step is to simply differentiate the equation with respect to t. 2A r rπ′ ′=


Calculus I

Step 3 Now, to finish this problem off we’ll first need to go back to the equation of the area and use the fact that we know the area at the point we are interested in and determine the radius at that time.

2 1212 1.9544r rππ

= ⇒ = =

The rate of change of the radius is then,

( )0.5 2 1.9544 0.040717r rπ ′ ′− = ⇒ = − 5. A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 ft/sec. At what rate is the distance between the person and the rocket increasing (a) 20 seconds after liftoff? (b) 1 minute after liftoff? Step 1 Here is a sketch for this situation that will work for both parts so we’ll put it here.

Step 2 In both parts we know that 15y′ = and want to determine z′ for each given time. Using the Pythagorean Theorem we get the following equation to relate y and z. 2 2 2 2350 122500z y y= + = + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 y yz z y y zz′

′ ′ ′= ⇒ =

We have now reached a point where the process will differ for each part.


Calculus I

(a) At what rate is the distance between the person and the rocket increasing 20 seconds after liftoff? To finish off this problem all we need to do is determine y (from the speed of the rocket and given time) and z (reusing the Pythagorean Theorem).

( )( ) 2 215 20 300 300 350 212500 50 85 460.9772y z= = = + = = =

The rate of change of the distance between the two is then,

( )( )300 159.76187

460.9772z′ = =

(b) At what rate is the distance between the person and the rocket increasing 1 minute after liftoff? This part is nearly identical to the first part with the exception that the time is now 60 seconds (and note that we MUST be in seconds because the speeds are in time of seconds). Here is the work for this problem.

( )( )

( )( )

2 215 60 900 900 350 932500 50 373 965.6604

900 1513.98007

965.6604

y z

z

= = = + = = =

′ = =

6. A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s and is initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar station? See the (probably bad) sketch below to help visualize the problem.

(a) At what rate is the distance between the plane and the radar station changing initially? Step 1


Calculus I

For this part we know that 100x′ = − when 2500x = . In this case note that x′ must be negative because x will be decreasing in this part. Also note that we converted x to meters since all the other quantities are in meters. Here is a sketch for this part.

Step 2 We want to determine z′ in this part so using the Pythagorean Theorem we get the following equation to relate x and z. 2 2 2 2750 562500z x x= + = + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 x xz z x x zz′

′ ′ ′= ⇒ =

Step 4 To finish off this problem all we need to do is determine z (reusing the Pythagorean Theorem) and then plug into the equation from Step 3 above.

2 22500 750 6812500 250 109 2610.0766z = + = = = The rate of change of the distance between the two for this part is,

( )( )2500 10095.7826

2610.0766z

−′ = = −

(b) At what rate is the distance between the plane and the radar station changing 30 seconds after it passes over the radar station? Step 1 For this part we know that 100x′ = and it will be positive in this case because x will now be increasing as we can see in the sketch below.


Calculus I

Step 2 As with the previous part we want to determine z′ and equation we’ll need is identical to the previous part so we’ll just rewrite both it and its derivative here.

2 2 2 2750 562500

2 2

z x xx xz z x x zz

= + = +′

′ ′ ′= ⇒ =

Step 3 To finish off this problem all we need to do is determine both x and z. For x we know the speed of the plane and the fact that it has flown for 30 seconds after passing over the radar station. So x is, ( )( )100 30 3000x = = For z we just need to reuse the Pythagorean Theorem.

2 23000 750 9562500 750 17 3092.3292z = + = = = The rate of change of the distance between the two for this part is then,

( )( )3000 10097.0143

3092.3292z′ = =

7. Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later? Step 1 Here is a sketch for this part.


Calculus I

We want to determine z′ after 15 seconds given that 2x′ = , 7y′ = and assuming that they start at the same point. Step 2 Hopefully it’s clear that we’ll need the Pythagorean Theorem to solve this problem so here is that. 2 2 2z x y= + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 2 x x y yz z x x y y zz′ ′+′ ′ ′ ′= + ⇒ =

Step 4 To finish off this problem all we need to do is determine all three lengths of the triangle in the sketch above. We can find x and y using their speeds and time while we can find z by reusing the Pythagorean Theorem.

( )( ) ( )( )2 2

2 15 30 7 15 105

30 105 11925 15 53 109.2016

x y

z

= = = =

= + = = =

The rate of change of the distance between the two people is then,

( )( ) ( )( )30 2 105 77.2801

109.2016z

+′ = =

8. Two people on bikes are at the same place. One of the bikers starts riding directly north at a rate of 8 m/sec. Five seconds after the first biker started riding north the second starts to ride directly east at a rate of 5 m/sec. At what rate is the distance between the two riders increasing 20 seconds after the second person started riding?


Calculus I

Step 1 Here is a sketch of this situation.

We want to determine z′ after 20 seconds after the second biker starts riding east given that

5x′ = , 8y′ = and assuming that they start at the same point. Step 2 Hopefully it’s clear that we’ll need the Pythagorean Theorem to solve this problem so here is that. 2 2 2z x y= + Step 3 Finally, let’s differentiate this with respect to t and we can even solve it for z′ so the actual solution will be quick and simple to find.

2 2 2 x x y yz z x x y y zz′ ′+′ ′ ′ ′= + ⇒ =

Step 4 To finish off this problem all we need to do is determine all three lengths of the triangle in the sketch above. We can find x and y using their speeds and time while we can find z by reusing the Pythagorean Theorem. Note that the biker riding east will be riding for 20 seconds and the biker riding north will be riding for 25 seconds (this biker started 5 seconds earlier…).

( )( ) ( )( )2 2

5 20 100 8 25 200

100 200 50000 100 5 223.6068

x y

z

= = = =

= + = = =

The rate of change of the distance between the two people is then,

( )( ) ( )( )100 5 200 89.3915

223.6068z

+′ = =

9. A light is mounted on a wall 5 meters above the ground. A 2 meter tall person is initially 10 meters from the wall and is moving towards the wall at a rate of 0.5 m/sec. After 4 seconds of moving is the tip of the shadow moving (a) towards or away from the person and (b) towards or away from the wall?


Calculus I

Step 1 Here is a sketch for this situation that will work for both parts so we’ll put it here. Also note that we know that 0.5px′ = − for both parts.

(a) After 4 seconds of moving is the tip of the shadow towards or away from the person? Step 2 In this case we want to determine sx′ when ( )10 4 0.5 8px = − = (although it will turn out that

we simply don’t need this piece of information for this problem….). We can use the idea of similar triangles to get the following equation.

25

s s

p s

x xx x x

= =+

If we solve this for sx we arrive at,

( )2

5

2 2 25 5 3 p

p s s

p s s s

x x x

x x x x x

+ =

+ = ⇒ =

This equation will work perfectly for us. Step 3 Differentiation with respect to t will give us, 2

3 psx x′ ′= Step 4 Finishing off this problem is very simple as all we need to do is plug in the known speed.

( )2 13 30.5sx′ = − = −


Calculus I

Because this rate is negative we can see that the tip of the shadow is moving towards the person at a rate of 1

3 m/s. (b) After 4 seconds of moving is the tip of the shadow towards or away from the wall? Step 2 In this case we want to determine x′ and the equation is really simple. All we need is, p sx x x= + Step 3 Differentiation with respect to t will give us, p sx x x′ ′ ′= + Step 4 Finishing off this problem is very simple as all we need to do is plug in the known speeds and note that we will need to result from the first part here. So we have 1

2px′ = − from the problem

statement and 13sx′ = − from the previous part.

( ) 51 12 3 6x′ = − + − = −

Because this rate is negative we can see that the tip of the shadow is moving towards the wall at a rate of 5

6 m/s. 10. A tank of water in the shape of a cone is being filled with water at a rate of 12 m3/sec. The base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the depth of the water in the tank changing with the radius of the top of the water is 10 meters? Step 1 Here is a sketch of the cross section of the tank and it is not even remotely to scale as I found it easier to reuse an old image that I had lying around. I can be a little lazy sometimes….


Calculus I

We want to determine h′ when 10r = and we know that 12V ′ = . Step 2 We’ll need the equation for the volume of a cone. 21

3V r hπ= This is a problem however as it has both r and h in it and it would be best to have only h since we need h′ . We can use similar triangles to fix this up. Based on similar triangles we get the following equation which can be solved for r.

134

268

r r hh= ⇒ =

Plugging this into the volume equation gives, 3169

48V hπ= Step 3 Next, let’s differentiate this with respect to t. 2169

16V h hπ′ ′= Step 4 To finish off this problem all we need to do is determine the value of h for the time we are interested in. This can easily be done from the similar triangle equation and the fact that we know 10r = .

( ) 404 413 13 1310h r= = =

The rate of change of the height of the water is then,


Calculus I

( )2169 40 316 13 2512 100h h h ππ π′ ′ ′= = ⇒ =

11. The angle of elevation is the angle formed by a horizontal line and a line joining the observer’s eye to an object above the horizontal line. A person is 500 feet way from the launch point of a hot air balloon. The hot air balloon is starting to come back down at a rate of 15 ft/sec. At what rate is the angle of elevation, θ , changing when the hot air balloon is 200 feet above the ground. See the (probably bad) sketch below to help visualize the angle of elevation if you are having trouble seeing it.

Step 1 Putting variables and known quantities on the sketch from the problem statement gives,

We want to determine θ ′ when 200y = and we know that 15y′ = − . Step 2 There are a variety of equations that we could use here but probably the best one that involves all of the known and needed quantities is,

( )tan500

yθ =

Step 3 Differentiating with respect to t gives,

( ) ( )2 2sec cos500 500y yθ θ θ θ′ ′

′ ′= ⇒ =

Step 4


Calculus I

To finish off this problem all we need to do is determine the value of θ for the time in question. We can either use the original equation to do this or we could acknowledge that all we really need is ( )cos θ and we could do a little right triangle trig to determine that.

For this problem we’ll just use the original equation to find the value of θ .

( ) ( )1200 2500 5tan tan 0.38051radiansθ θ −= ⇒ = =

The rate of change of the angle of elevation is then,

( )215 cos 0.38051 0.02586500

θ −′ = = −

Higher Order Derivatives 1. Determine the fourth derivative of ( ) 7 4 33 6 8 12 18h t t t t t= − + − +

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then, ( ) 6 3 221 24 24 12h t t t t′ = − + − Step 2 The second derivative is, ( ) 5 2126 72 48h t t t t′′ = − +

Step 3 The third derivative is, ( ) 4630 144 48h t t t′′′ = − + Step 4 The fourth, and final derivative for this problem, is,

( ) ( )4 32520 144h t t= −


Calculus I

2. Determine the fourth derivative of ( ) 3 2 1V x x x x= − + −

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then, ( ) 23 2 1V x x x′ = − + Step 2 The second derivative is, ( ) 6 2V x x′′ = −

Step 3 The third derivative is, ( ) 6V x′′′ = Step 4 The fourth, and final derivative for this problem, is,

( ) ( )4 0V x = Note that we could have just as easily used the Fact from the notes to arrive at this answer in one step.

3. Determine the fourth derivative of ( ) 5 32

148

f x x xx

= − −

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. After a quick rewrite of the function to help with the differentiation the first derivative is,

( ) ( )3 21 1

2 35 52 21 12 1 148 5 4 2

f x x x x f x x x x− −− −′= − − ⇒ = + −

Step 2 The second derivative is,

( )7 3

45 224 3 125 4 4

f x x x x− −−′′ = − − +

Step 3 The third derivative is,


Calculus I

( )12 5

55 2168 33125 8

f x x x x− −−′′′ = + −

Step 4 The fourth, and final derivative for this problem, is,

( ) ( )17 7

4 65 22016 1515625 16

f x x x x− −−= − − +

4. Determine the fourth derivative of ( ) ( ) ( )37sin cos 1 2wf w w= + −

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then,

( ) ( ) ( )37 cos 2sin 1 23

wf w w′ = + −


( ) ( ) ( )37 sin 4cos 1 29

wf w w′′ = − − −


( ) ( ) ( )37 cos 8sin 1 227

wf w w′′′ = − − −


( ) ( ) ( ) ( )43

7 sin 16cos 1 281

wf w w= + −

5. Determine the fourth derivative of ( )5 48ln 2zy z−= +e

Step 1 Not much to this problem other than to take four derivatives so each step will show each successive derivatives until we get to the fourth. The first derivative is then,


Calculus I

3

5 5 14

85 8 5 322

z zdy z zdx z

− − − = − + = − +

e e


2

5 22 25 32zd y z

dx− −= −e


3

5 33 125 64zd y z

dx− −= − +e


4

5 44 625 192zd y z

dx− −= −e

6. Determine the second derivative of ( ) ( )3sin 2 9g x x x= −

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. The first derivative is then,

( ) ( ) ( )2 36 9 cos 2 9g x x x x′ = − − Step 2 Do not forget that often we will end up needing to do a product rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,

( ) ( ) ( ) ( )23 2 312 cos 2 9 6 9 sin 2 9g x x x x x x x′′ = − − − −

7. Determine the second derivative of ( )3ln 7z x= −

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. The first derivative is then,


Calculus I

2

3

37

dz xdx x

−=

−

Step 2 Do not forget that often we will end up needing to do a quotient rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,

( ) ( )( )

( ) ( )

3 2 22 4

2 22 3 3

6 7 3 3 42 3

7 7

x x x xd z x xdx x x

− − − − − − −= =

− −

8. Determine the second derivative of ( )( )42

2

6 2Q v

v v=

+ −

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. We’ll do a quick rewrite of the function to help with the derivatives and then the first derivative is,

( ) ( )( ) ( )( )

42

52

2 6 2

8 2 2 6 2

Q v v v

Q v v v v

−

−

= + −

′ = − − + −

Step 2 Do not forget that often we will end up needing to do a product rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,

( ) ( ) ( ) ( )5 622 216 6 2 40 2 2 6 2Q v v v v v v− −

′′ = + − + − + − 9. Determine the second derivative of ( ) ( )2cos 7H t t=

Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. The first derivative is then, ( ) ( ) ( )14cos 7 sin 7H t t t′ = − Step 2 Do not forget that often we will end up needing to do a product rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,


Calculus I

( ) ( ) ( ) ( ) ( ) ( ) ( )2 298sin 7 sin 7 98cos 7 cos 7 98sin 7 98cos 7H t t t t t t t′′ = − = − Note that, in this case, if we recall our trig formulas we could have reduced the product in the first derivative to a single trig function which would have then allowed us to avoid the product rule for the second derivative. Can you figure out what the formula is? 10. Determine the second derivative of 3 22 1 4x y y+ = − Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. Note however that we are going to have to do implicit differentiation to do each derivative. Here is the work for the first derivative. If you need a refresher on implicit differentiation go back to that section and check some of the problems in that section.

( )

2

2 22

6 2 46 32 4 6

2 4 2

x y y yx xy y x y

y y

′ ′+ = −

− −′ ′+ = − ⇒ = =+ +

Step 2 Now, the second derivative will also need implicit differentiation. Note as well that we can work with the first derivative in its present form which will require the quotient rule or we can rewrite it as,

( ) 123 2y x y −′ = − + and use the product rule. These get messy enough as it is so we’ll go with the product rule to try and keep the “mess” down a little. Using implicit differentiation to take the derivative of first derivative gives,

( ) ( ) ( )1 226 2 3 2dy y x y x y ydx

− −′′ ′ ′= = − + + + Step 3 Finally, recall that we don’t want a y′ in the second derivative so to finish this out we need to plug in the formula for y′ (which we know…) and do a little simplifying to get the final answer.

( ) ( ) ( )( ) ( ) ( )1 2 1 1 32 2 46 2 3 2 3 2 6 2 9 2y x y x y x y x y x y− − − − −′′ = − + + + − + = − + − +


Calculus I

11. Determine the second derivative of 26 1y xy− = Step 1 Not much to this problem other than to take two derivatives so each step will show each successive derivatives until we get to the second. Note however that we are going to have to do implicit differentiation to do each derivative. Here is the work for the first derivative. If you need a refresher on implicit differentiation go back to that section and check some of the problems in that section.

( )

2

22

6 2 0

6 26 2

y y xy yyxy y y y

xy

′ ′− − =

′ ′− = ⇒ =−

Step 2 Now, the second derivative will also need implicit differentiation. Note as well that we can work with the first derivative in its present form which will require the quotient rule or we can rewrite it as,

( ) 12 6 2y y xy −′ = − and use the product rule. These get messy enough as it is so we’ll go with the product rule to try and keep the “mess” down a little. Using implicit differentiation to take the derivative of first derivative gives,

( ) ( ) ( ) ( )1 222 6 2 6 2 2 2dy y y y xy y xy y xydx

− −′′ ′ ′ ′= = − − − − − Step 3 Finally, recall that we don’t want a y′ in the second derivative. So, to finish this out let’s do a little “simplifying” of the to make it “easier” to plug in the formula for y′ .

( ) ( ) ( )( ) ( )( ) ( )

1 2 23 2

1 1 23

2 6 2 2 6 2 2 6 2

2 6 2 1 6 2 2 6 2

y y y xy y xy xy y xy

y y xy xy xy y xy

− − −

− − −

′′ ′ ′= − + − + −

′= − + − + −

The point of all of this was to get down to a single y′ in the formula for the second derivative, which won’t always be possible to do, and a little factoring to try and make things a little easier to deal with. Finally all we need to do is plug in the formula for y′ to get the final answer.


Calculus I

( ) ( ) ( )( ) ( )

( ) ( )( ) ( )

1 1 1 22 3

2 1 23 3

2 6 2 6 2 1 6 2 2 6 2

2 6 2 1 6 2 2 6 2

y y y xy xy xy xy y xy

y xy xy xy y xy

− − − −

− − −

′′ = − − + − + −

= − + − + −

Note that for a further simplification step, if we wanted to go further, we could factor a

( ) 232 6 2y xy −− out of both terms to get,

( ) ( )( )2 132 6 2 2 6 2y y xy xy xy− −′′ = − + −

Logarithmic Differentiation 1. Use logarithmic differentiation to find the first derivative of

( ) ( )72 25 3 6 8 12f x x x x= − + − .

Step 1 Take the logarithm of both sides and do a little simplifying.

( ) ( )

( ) ( )( ) ( )

12

72 2

72 2

2 212

ln ln 5 3 6 8 12

ln 5 3 ln 6 8 12

7 ln 5 3 ln 6 8 12

f x x x x

x x x

x x x

= − + − = − + + −

= − + + −

Step 2 Use implicit differentiation to differentiate both sides with respect to x.

( )( ) 2 2 2 2

6 1 12 8 42 6 475 3 2 6 8 12 5 3 6 8 12

f x x x x xf x x x x x x x′ − + − +

= + = +− + − − + −

Step 3 Finally, solve for the derivative and plug in the equation for ( )f x .


Calculus I

( ) ( )

( )

2 2

72 22 2

42 6 45 3 6 8 12

42 6 45 3 6 8 125 3 6 8 12

x xf x f xx x x

x xx x xx x x

− + ′ = + − + −

− + = − + − + − + −

2. Use logarithmic differentiation to find the first derivative of ( )

( )

2

34

sin 3

6

z zy

z

+=

−.


( ) ( )

( )( ) ( )

( )

232 4

34

2 4

sin 3ln ln ln sin 3 ln 6

6

ln sin 3 3ln 6

z zy z z z

z

z z z

+ = = + − − − = + − −

Step 2 Use implicit differentiation to differentiate both sides with respect to z.

( ) ( )

( ) ( ) ( )2 3 3

24 42

3 2 cos 3 4 123 3 2 cot 36 6sin 3

z z zy z zz z zy z zz z

+ +′ −= − = + + + − −+

Step 3 Finally, solve for the derivative and plug in the equation for y .

( ) ( ) ( )( )

( ) ( )23 3

2 234 44

sin 312 123 2 cot 3 3 2 cot 36 66

z zz zy y z z z z z zz zz

+ ′ = + + + = + + + − − −

3. Use logarithmic differentiation to find the first derivative of ( ) ( )3

24

5 8 1 9cos 4

10

t th t

t t

+ −=

+.



Calculus I

( ) ( )

( )

( ) ( )( ) ( )

( ) ( )( ) ( )

3

24

243

1112 432

21 1 12 3 4

5 8 1 9cos 4ln ln

10

ln 5 8 1 9cos 4 ln 10

ln 5 8 ln 1 9cos 4 ln 10

ln 5 8 ln 1 9cos 4 10

t th t

t t

t t t t

t t t t

t t t t

+ − = +

= + − − + = + + − − +

= + + − − +

Note that the logarithm simplification work was a little complicated for this problem, but if you know your logarithm properties you should be okay with that. Step 2 Use implicit differentiation to differentiate both sides with respect to t.

( )( )

( )( )

1 1 12 3 4 2

36sin 45 2 105 8 1 9cos 4 10

h t t th t t t t t′ +

= + −+ − +

Step 3 Finally, solve for the derivative and plug in the equation for ( )h t .

( ) ( ) ( )( )

( ) ( )( )

5 512 2 2

2

3 5 512 2 2

224

12sin 45 8 1 9cos 4 10

5 8 1 9cos 4 12sin 45 8 1 9cos 4 1010

t th t h tt t t t

t t t tt t t tt t

+′ = + − + − +

+ − += + − + − ++

4. Find the first derivative of ( ) ( )43 7 wg w w= − .

Step 1 We just need to do some logarithmic differentiation so take the logarithm of both sides and do a little simplifying.

( ) ( ) ( )4ln ln 3 7 4 ln 3 7wg w w w w = − = −

Step 2 Use implicit differentiation to differentiate both sides with respect to w. Don’t forget to product rule the right side.

( )( ) ( ) ( )3 124ln 3 7 4 4ln 3 7

3 7 3 7g w ww w wg w w w′

= − + = − +− −


Calculus I

Step 3 Finally, solve for the derivative and plug in the equation for ( )g w .

( ) ( ) ( )

( ) ( )4

124ln 3 73 7

123 7 4ln 3 73 7

w

wg w g w ww

ww ww

′ = − + −

= − − + −

5. Find the first derivative of ( ) ( ) ( )sin 282xxf x x= − e .

Step 1 We just need to do some logarithmic differentiation so take the logarithm of both sides and do a little simplifying.

( ) ( ) ( ) ( ) ( )sin 28 8ln ln 2 sin 2 ln 2xx xf x x x x = − = −

e e

Step 2 Use implicit differentiation to differentiate both sides with respect to x. Don’t forget to product rule the right side.

( )( ) ( ) ( ) ( )

( ) ( ) ( )

88

8

88

8

2 82cos 2 ln 2 sin 22

2 82cos 2 ln 2 sin 22

xx

x

xx

x

f xx x x

f x x

x x xx

′ −= − +

−

−= − +

−

eee

eee

Step 3 Finally, solve for the derivative and plug in the equation for ( )f x .

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

88

8

8sin 28 88

2 82cos 2 ln 2 sin 22

2 82 2cos 2 ln 2 sin 22

x

xx

x

xx x

x

f x f x x x xx

x x x xx

−′ = − + −

−= − − + −

eee

ee ee


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